Integration

Class 12 Mathematics, unit 3, book pages 89–119

Integration is the reverse of differentiation: it finds a function from its derivative. This unit builds the table of standard integrals and three main techniques (substitution, integration by parts and partial fractions), then defines the definite integral as a limit of sums and evaluates it with the Fundamental Theorem of Calculus. Definite integrals then give areas under and between curves, volumes of solids of revolution, average values, and applications such as consumer and producer surplus, motion and the work done in stretching a spring.

Learning outcomes (18)

Find the general antiderivative of a given function.
Recognise and use the terms and notations for antiderivatives.
State the power rule of integrals.
State and apply the properties of indefinite integrals.
Integrate functions involving the exponential and logarithmic functions.
Identify when to use integration by parts to solve integration problems.
Apply the integration by parts formula for definite integrals.
Solve integration problems involving trigonometric substitution.
Integrate a rational function using the method of partial fractions.
State the definition of the definite integral.
Explain the terms integrand, limits of integration and value of integration.
State and apply the properties of definite integrals.
State and apply the fundamental theorem of calculus to evaluate definite integrals.
Describe the relation between the definite integral and net area.
Find the area of a region bounded by a curve and lines parallel to the axes, or between a curve and a line, or between two curves.
Find the volume of revolution about one of the axes.
Demonstrate the trapezium rule to estimate the value of a definite integral.
Apply integration to real-world problems such as the volume of a container, consumer and producer surplus, the growth rate of a population, the time period for an investment return, and the drug dosage required, by integrating the concentration.

3.1Integration and antiderivativesOften examined

You already know how to differentiate: from a function $F (x)$ you find its derivative $F^{'} (x)$ . Integration does the opposite. We are given the derivative $f (x)$ and we look for the function it came from. Subtraction undoes addition and division undoes multiplication; in the same way integration (also called anti-differentiation) undoes differentiation.

The unit has two main parts. First we find functions from their derivatives (indefinite integrals). Then we find areas and volumes by adding up very many very thin pieces (definite integrals). The Fundamental Theorem of Calculus joins the two ideas. Newton and Leibniz discovered this area of mathematics independently.

Antiderivative (Definition 3.1): A function $F (x)$ is an antiderivative of $f (x)$ on an interval if $F^{'} (x) = f (x)$ for every $x$ in that interval.

\frac{d}{d x} (\frac{x ^{4}}{4}) = x^{3}, \frac{d}{d x} (\frac{x ^{4}}{4} + 3) = x^{3}, \frac{d}{d x} (\frac{x ^{4}}{4} - π) = x^{3}

So $\frac{1}{4} x^{4}$ , $\frac{1}{4} x^{4} + 3$ , $\frac{1}{4} x^{4} - π$ and $\frac{1}{4} x^{4} + c$ (with $c$ any constant) are all antiderivatives of $x^{3}$ . A function has infinitely many antiderivatives. If $F (x)$ is one of them, then $F (x) + c$ is another, because the derivative of a constant is zero:

\frac{d}{d x} [F (x) + c] = \frac{d}{d x} [F (x)] + \frac{d}{d x} [c] = f (x) + 0 = f (x)

We write this whole family of antiderivatives with the integral sign:

\int f (x) d x = F (x) + c

Read it as 'the integral of $f (x)$ with respect to $x$ equals $F (x)$ plus $c$ '.

Integral sign $\int$: The symbol that tells us to integrate. It is a stretched letter S, for 'sum'.
Integrand: The function $f (x)$ that is being integrated.
$d x$: Shows that we integrate with respect to $x$ , the variable of integration. Other letters can be used: $\int y^{2} d y$ , $\int sin θ d θ$ .
Constant of integration: The arbitrary constant $c$ added to every indefinite integral.
Indefinite integral: The family $F (x) + c$ of all antiderivatives of $f (x)$ . It is called indefinite because $c$ is not fixed.

Reading a derivative backwards

Use

\frac{d}{d x} (x^{4}) = 4 x^{3}

to write down

\int 4 x^{3} d x

, and list some antiderivatives of

4 x^{3}

The derivative of $x^{4}$ is $4 x^{3}$ , so $x^{4}$ is an antiderivative of $4 x^{3}$ . Add the constant of integration.
$\int 4 x^{3} d x = x^{4} + c$
Each value of $c$ gives one antiderivative.
$x^{4}, x^{4} + 1, x^{4} - 2, x^{4} + π are all antiderivatives of 4 x^{3}$

Three antiderivatives of

f (x) = 2 x

−3

−2

−1

−3

−2

−1

x

y

y = x^{2} + 2

y = x^{2}

y = x^{2} - 2

Every antiderivative of

2 x

has the form

x^{2} + c

. The curves are vertical shifts of one another. At

x = 1

all of them have the same slope,

2

(dotted tangent lines), because they all have the same derivative

2 x

Using a known value to find c

Find

f (x)

f^{'} (x) = 3 x^{2}

and

f (1) = 5

Find all antiderivatives of $3 x^{2}$ . The derivative of $x^{3}$ is $3 x^{2}$ .
$f (x) = \int 3 x^{2} d x = x^{3} + c$
Use the known value $f (1) = 5$ to find $c$ .
$5 = 1^{3} + c ⟹ c = 4$

f (x) = x^{3} + 4

3.1Basic integration formulasOften examined

Each basic formula comes from a derivative formula read backwards. The method is always the same: differentiate a suitable function, then integrate both sides. Integrating a derivative gives back the original function (plus a constant).

Derivation Formula 3.1: the power rule

Aim. Show that $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ for $n \neq = - 1$ .

Differentiate $\frac{x ^{n + 1}}{n + 1} + c$ . The power rule for derivatives brings down the factor $n + 1$ , which cancels with the denominator. The derivative of $c$ is $0$ .
$\frac{d}{d x} [\frac{x ^{n + 1}}{n + 1} + c] = \frac{( n + 1 ) x ^{n}}{n + 1} + 0 = x^{n} (i)$
Integrate both sides of (i) with respect to $x$ . On the left, integration undoes the differentiation.
$\int \frac{d}{d x} [\frac{x ^{n + 1}}{n + 1} + c] d x = \int x^{n} d x ⟹ \frac{x ^{n + 1}}{n + 1} + c = \int x^{n} d x$
We divided by $n + 1$ , so the formula fails when $n + 1 = 0$ .
$n + 1 \neq = 0 ⟺ n \neq = - 1$

Result

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c, n \neq = - 1

To integrate a power of $x$ , add 1 to the power and divide by the new power. The case $n = - 1$ , that is $\int x^{- 1} d x$ , needs its own formula (Formula 3.2).

In general, if a bracket is raised to a power and the derivative of the bracket is also a factor of the integrand:

\int [f (x)]^{n} f^{'} (x) d x = \frac{[ f ( x ) ] ^{n + 1}}{n + 1} + c, n \neq = - 1

General power rule

Evaluate

\int 2 x (x^{2} + 1)^{4} d x

The bracket is $f (x) = x^{2} + 1$ , and its derivative $f^{'} (x) = 2 x$ is present as a factor.
$\int (x^{2} + 1)^{4} (2 x) d x = \int [f (x)]^{4} f^{'} (x) d x$
Apply the general power rule with $n = 4$ .
$= \frac{( x ^{2} + 1 ) ^{5}}{5} + c$

\frac{( x ^{2} + 1 ) ^{5}}{5} + c

Derivation Formula 3.2: the integral of 1/x

Aim. Show that $\int \frac{1}{x} d x = ln ∣ x ∣ + c$ .

For $x > 0$ , the derivative of $ln x$ is $\frac{1}{x}$ .
$\frac{d}{d x} [ln x + c] = \frac{1}{x} (ii)$
For $x < 0$ , $ln x$ is not defined but $ln (- x)$ is, and by the chain rule its derivative is also $\frac{1}{x}$ . So $\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}$ for every $x \neq = 0$ .
$\frac{d}{d x} ln (- x) = \frac{- 1}{- x} = \frac{1}{x}$
Integrate both sides with respect to $x$ .
$\int \frac{d}{d x} [ln ∣ x ∣ + c] d x = \int \frac{1}{x} d x ⟹ ln ∣ x ∣ + c = \int \frac{1}{x} d x$

Result

\int \frac{1}{x} d x = ln ∣ x ∣ + c, x \neq = 0

This fills the gap $n = - 1$ in the power rule. The book writes $ln x + c$ , which is correct for $x > 0$ ; the absolute value makes the formula true for negative $x$ as well.

In general, if the numerator is the derivative of the denominator, the integral is the natural log of the denominator:

\int \frac{f ^{'} ( x )}{f ( x )} d x = ln f (x) + c

Derivation Formula 3.3: the integral of e^x

Aim. Show that $\int e^{x} d x = e^{x} + c$ .

The exponential function is its own derivative.
$\frac{d}{d x} [e^{x} + c] = e^{x} (iii)$
Integrate both sides of (iii) with respect to $x$ .
$\int \frac{d}{d x} [e^{x} + c] d x = \int e^{x} d x$
The left side is the function we differentiated.
$e^{x} + c = \int e^{x} d x$

Result

\int e^{x} d x = e^{x} + c

$e^{x}$ is unchanged by differentiation and by integration (apart from the constant).

\int e^{f (x)} f^{'} (x) d x = e^{f (x)} + c

For example $\int e^{s i n x} cos x d x = e^{s i n x} + c$ , because $cos x$ is the derivative of the power $sin x$ .

Derivation Formula 3.4: the integral of a^x

Aim. Show that $\int a^{x} d x = \frac{a ^{x}}{ln a} + c$ for $a > 0$ , $a \neq = 1$ .

The derivative of $a^{x}$ is $a^{x} ln a$ . Here $ln a$ is a constant, so dividing by it leaves exactly $a^{x}$ .
$\frac{d}{d x} [\frac{1}{ln a} a^{x} + c] = \frac{1}{ln a} a^{x} ln a = a^{x} (iv)$
Integrate both sides of (iv) with respect to $x$ .
$\int \frac{d}{d x} [\frac{a ^{x}}{ln a} + c] d x = \int a^{x} d x$
The left side is the function we differentiated. We need $a \neq = 1$ because $ln 1 = 0$ .
$\frac{a ^{x}}{ln a} + c = \int a^{x} d x$

Result

\int a^{x} d x = \frac{a ^{x}}{ln a} + c, a > 0, a \neq = 1

An exponential with any positive base integrates to itself divided by the natural log of the base. With $a = e$ , $ln e = 1$ and Formula 3.3 comes back.

\int a^{f (x)} f^{'} (x) d x = \frac{a ^{f (x)}}{ln a} + c

When the inside is a linear function ax + b

If $x$ is replaced by $a x + b$ (with $a \neq = 0$ ), integrate in the usual way and then divide by $a$ , the coefficient of $x$ . Differentiating the answer brings back the factor $a$ by the chain rule, and dividing by $a$ cancels it.

Standard results for a linear inner function. For example $\int e^{3 x} d x = \frac{1}{3} e^{3 x} + c$ and $\int (2 x + 1)^{3} d x = \frac{1}{8} (2 x + 1)^{4} + c$ .
Integrand	Integral
$(a x + b)^{n}$ , $n \neq = - 1$	$\frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + c$
$\frac{1}{a x + b}$	$\frac{1}{a} ln ∣ a x + b ∣ + c$
$e^{a x + b}$	$\frac{1}{a} e^{a x + b} + c$
$k^{a x + b}$ ( $k > 0$ , $k \neq = 1$ )	$\frac{k ^{a x + b}}{a ln k} + c$

3.1Properties of indefinite integralsOften examined

These rules let us break a long integrand into simple pieces and integrate term by term. They follow from the matching rules for derivatives.

Theorem 3.1

(i) A constant factor can be moved through the integral sign: \int c\,f(x)\,dx=c\int f(x)\,dx.
(ii) An antiderivative of a sum is the sum of the antiderivatives: \int[f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx.
(iii) An antiderivative of a difference is the difference of the antiderivatives: \int[f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx.
(iv) In general: \int[a f(x)\pm b g(x)]\,dx=a\int f(x)\,dx\pm b\int g(x)\,dx.

Example 1

Evaluate (i) \int(4x^{7}-2x^{3}+9x+3)\,dx (ii) \displaystyle\int\frac{y^{3}-2y^{6}}{y^{5}}\,dy.

(i)

Split the integral term by term and take the constants outside (Theorem 3.1).
\int(4x^{7}-2x^{3}+9x+3)\,dx=4\int x^{7}dx-2\int x^{3}dx+9\int x\,dx+3\int dx
Apply the power rule to each term. Only one constant c is needed, at the end.
=4\left(\dfrac{x^{8}}{8}\right)-2\left(\dfrac{x^{4}}{4}\right)+9\left(\dfrac{x^{2}}{2}\right)+3x+c=\dfrac{x^{8}}{2}-\dfrac{x^{4}}{2}+\dfrac{9x^{2}}{2}+3x+c

(ii)

Divide each term of the numerator by y^{5} so that every term is a single power of y.
\dfrac{y^{3}-2y^{6}}{y^{5}}=\dfrac{y^{3}}{y^{5}}-\dfrac{2y^{6}}{y^{5}}=y^{-2}-2y
Integrate with the power rule.
\int(y^{-2}-2y)\,dy=\dfrac{y^{-1}}{-1}-2\left(\dfrac{y^{2}}{2}\right)+c=-\dfrac{1}{y}-y^{2}+c

Example 2

Evaluate (i) \displaystyle\int\frac{ax+\frac12 b}{ax^{2}+bx+c}\,dx (ii) \int e^{3x}\,dx.

(i)

The derivative of the denominator ax^{2}+bx+c is 2ax+b. The numerator ax+\tfrac12b is half of this, so multiply and divide by 2.
\int\dfrac{ax+\frac12b}{ax^{2}+bx+c}\,dx=\dfrac12\int\dfrac{2ax+b}{ax^{2}+bx+c}\,dx
Now the numerator is the derivative of the denominator, so use \int\dfrac{f'(x)}{f(x)}dx=\ln|f(x)|. The constant is called C here, because c is already used in the question.
=\dfrac12\ln\left|ax^{2}+bx+c\right|+C

(ii)

The derivative of the power 3x is 3. Multiply and divide by 3, then use \int e^{f(x)}f'(x)\,dx=e^{f(x)}+c.
\int e^{3x}\,dx=\dfrac13\int e^{3x}(3)\,dx=\dfrac13e^{3x}+c

Example 3

Evaluate \displaystyle\int\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\,dx.

The power of e is f(x)=\sin^{-1}x, and its derivative is the other factor of the integrand.
f(x)=\sin^{-1}x\;\Longrightarrow\;f'(x)=\dfrac{1}{\sqrt{1-x^{2}}}
So the integrand is e^{f(x)}f'(x). Use \int e^{f(x)}f'(x)\,dx=e^{f(x)}+c.
\int e^{\sin^{-1}x}\cdot\dfrac{1}{\sqrt{1-x^{2}}}\,dx=e^{\sin^{-1}x}+c

e^{\sin^{-1}x}+c

3.2Integration of trigonometric functionsOften examined

Every derivative formula of a trigonometric function gives an integral formula when it is read backwards. Angles are always measured in radians.

Basic trigonometric integrals. When the angle is kx instead of x, divide by k.
Derivative	Integral
\dfrac{d}{dx}(\sin x)=\cos x	\int\cos x\,dx=\sin x+c
\dfrac{d}{dx}(\cos x)=-\sin x	\int\sin x\,dx=-\cos x+c
\dfrac{d}{dx}(\sin kx)=k\cos kx	\int\cos kx\,dx=\dfrac{\sin kx}{k}+c
\dfrac{d}{dx}(\cos kx)=-k\sin kx	\int\sin kx\,dx=-\dfrac{\cos kx}{k}+c
\dfrac{d}{dx}(\tan x)=\sec^{2}x	\int\sec^{2}x\,dx=\tan x+c
\dfrac{d}{dx}(\cot x)=-\operatorname{cosec}^{2}x	\int\operatorname{cosec}^{2}x\,dx=-\cot x+c
\dfrac{d}{dx}(\sec x)=\sec x\tan x	\int\sec x\tan x\,dx=\sec x+c
\dfrac{d}{dx}(\operatorname{cosec} x)=-\operatorname{cosec} x\cot x	\int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x+c

Example 4

Evaluate (i) \int\sin2x\,dx (ii) \int\cos\frac{3x}{5}\,dx (iii) \int\sec^{2}mx\,dx (iv) \int5\operatorname{cosec}^{2}\frac{7x}{5}\,dx (v) \int9\sec3x\tan3x\,dx.

(i) Here k=2. Multiply and divide by 2 so that the derivative of 2x appears.
\int\sin2x\,dx=\dfrac12\int\sin2x\,(2)\,dx=\dfrac12(-\cos2x)+c=-\dfrac{\cos2x}{2}+c
(ii) Here k=\tfrac35. Multiply by \tfrac35 inside and by \tfrac53 outside.
\int\cos\dfrac{3x}{5}\,dx=\dfrac53\int\cos\dfrac{3x}{5}\left(\dfrac35\right)dx=\dfrac53\sin\dfrac{3x}{5}+c
(iii) Here k=m, a non-zero constant.
\int\sec^{2}mx\,dx=\dfrac1m\int\sec^{2}mx\,(m)\,dx=\dfrac1m\tan mx+c
(iv) Take out 5; then k=\tfrac75.
\int5\operatorname{cosec}^{2}\dfrac{7x}{5}\,dx=5\times\dfrac57\int\operatorname{cosec}^{2}\dfrac{7x}{5}\left(\dfrac75\right)dx=\dfrac{25}{7}\left(-\cot\dfrac{7x}{5}\right)+c=-\dfrac{25}{7}\cot\dfrac{7x}{5}+c
(v) Write 9=3\times3; one factor 3 is the derivative of 3x.
\int9\sec3x\tan3x\,dx=3\int\sec3x\tan3x\,(3)\,dx=3\sec3x+c

Integrals of tan x, cot x, sec x and cosec x (Example 5)

These four do not come straight from the derivative table. Each is rearranged into the form \displaystyle\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c. They are standard 'prove that' questions.

Derivation Integral of tan x

Aim. Prove that \int\tan x\,dx=-\ln|\cos x|+c=\ln|\sec x|+c.

Write \tan x as sine over cosine.
\int\tan x\,dx=\int\dfrac{\sin x}{\cos x}\,dx
The derivative of the denominator \cos x is -\sin x. Put a minus sign inside and outside so that the numerator becomes this derivative.
=-\int\dfrac{-\sin x}{\cos x}\,dx=-\int\dfrac{\frac{d}{dx}(\cos x)}{\cos x}\,dx
Use \int\dfrac{f'}{f}dx=\ln|f|+c.
=-\ln|\cos x|+c
Use the log law -\ln A=\ln A^{-1} and \dfrac{1}{\cos x}=\sec x.
=\ln|\cos x|^{-1}+c=\ln\left|\dfrac{1}{\cos x}\right|+c=\ln|\sec x|+c

Result\int\tan x\,dx=-\ln|\cos x|+c=\ln|\sec x|+c

Both forms are correct; they are equal by the laws of logarithms.

Derivation Integral of cot x (check point)

Aim. Prove that \int\cot x\,dx=\ln|\sin x|+c.

Write \cot x as cosine over sine.
\int\cot x\,dx=\int\dfrac{\cos x}{\sin x}\,dx
The numerator \cos x is exactly the derivative of the denominator \sin x.
=\int\dfrac{\frac{d}{dx}(\sin x)}{\sin x}\,dx
Use \int\dfrac{f'}{f}dx=\ln|f|+c.
=\ln|\sin x|+c

Result\int\cot x\,dx=\ln|\sin x|+c

This is the check point of Section 3.2. The book writes \ln(\sin x); the absolute value keeps the formula valid where \sin x<0.

Derivation Integral of sec x

Aim. Prove that \int\sec x\,dx=\ln|\sec x+\tan x|+c.

Multiply and divide by (\sec x+\tan x). This does not change the integrand, but it creates a useful numerator.
\int\sec x\,dx=\int\dfrac{\sec x(\sec x+\tan x)}{\sec x+\tan x}\,dx=\int\dfrac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}\,dx
Differentiate the denominator: \dfrac{d}{dx}(\sec x+\tan x)=\sec x\tan x+\sec^{2}x. This is exactly the numerator.
=\int\dfrac{\frac{d}{dx}(\sec x+\tan x)}{\sec x+\tan x}\,dx
Use \int\dfrac{f'}{f}dx=\ln|f|+c.
=\ln|\sec x+\tan x|+c

Result\int\sec x\,dx=\ln|\sec x+\tan x|+c

Multiplying by \sec x+\tan x turns the integrand into 'derivative over function'. This result is used again in trigonometric substitution (Formulas 3.7 and 3.8).

Derivation Integral of cosec x

Aim. Prove that \int\operatorname{cosec} x\,dx=\ln|\operatorname{cosec} x-\cot x|+c.

Multiply and divide by (\operatorname{cosec} x-\cot x).
\int\operatorname{cosec} x\,dx=\int\dfrac{\operatorname{cosec} x(\operatorname{cosec} x-\cot x)}{\operatorname{cosec} x-\cot x}\,dx=\int\dfrac{\operatorname{cosec}^{2}x-\operatorname{cosec} x\cot x}{\operatorname{cosec} x-\cot x}\,dx
Differentiate the denominator: \dfrac{d}{dx}(\operatorname{cosec} x-\cot x)=-\operatorname{cosec} x\cot x+\operatorname{cosec}^{2}x. This is the numerator.
=\int\dfrac{\frac{d}{dx}(\operatorname{cosec} x-\cot x)}{\operatorname{cosec} x-\cot x}\,dx
Use \int\dfrac{f'}{f}dx=\ln|f|+c.
=\ln|\operatorname{cosec} x-\cot x|+c

Result\int\operatorname{cosec} x\,dx=\ln|\operatorname{cosec} x-\cot x|+c

The same idea as for \sec x, with \operatorname{cosec} x-\cot x as the multiplier.

Example 7 (i)

Integrate 8\sec9x-\tan3x with respect to x.

Split the integral and adjust each part so that the derivative of the angle appears (9 for 9x, 3 for 3x).
\int(8\sec9x-\tan3x)\,dx=\dfrac89\int\sec9x\,(9)\,dx-\dfrac13\int\tan3x\,(3)\,dx
Use \int\sec u\,du=\ln|\sec u+\tan u| and \int\tan u\,du=\ln|\sec u|.
=\dfrac89\ln|\sec9x+\tan9x|-\dfrac13\ln|\sec3x|+c

\dfrac{8}{9}\ln|\sec 9x+\tan 9x|-\dfrac{1}{3}\ln|\sec 3x|+c

3.2.1Integration of sin²x and cos²xOften examined

There is no basic formula for \int\sin^{2}x\,dx or \int\cos^{2}x\,dx. The trick is to use a double-angle identity to replace the square by a first power of \cos2x, which we can integrate.

Derivation The power-reducing identities

Aim. Express \sin^{2}x and \cos^{2}x in terms of \cos2x.

Start from two forms of the double-angle formula for cosine.
\cos2x=1-2\sin^{2}x\qquad\text{and}\qquad\cos2x=2\cos^{2}x-1
Make \sin^{2}x the subject of the first form.
2\sin^{2}x=1-\cos2x\;\Longrightarrow\;\sin^{2}x=\dfrac{1-\cos2x}{2}
Make \cos^{2}x the subject of the second form.
2\cos^{2}x=1+\cos2x\;\Longrightarrow\;\cos^{2}x=\dfrac{1+\cos2x}{2}

Result\sin^{2}x=\dfrac{1-\cos2x}{2},\qquad \cos^{2}x=\dfrac{1+\cos2x}{2}

The square of a sine or cosine becomes a constant plus a cosine of the double angle, and both parts are easy to integrate.

Example 6

Evaluate \int\sin^{2}x\,dx.

Replace \sin^{2}x using the identity and split the integral.
\int\sin^{2}x\,dx=\int\dfrac{1-\cos2x}{2}\,dx=\dfrac12\int dx-\dfrac12\int\cos2x\,dx
Integrate: \int\cos2x\,dx=\dfrac{\sin2x}{2}.
=\dfrac12x-\dfrac12\left(\dfrac{\sin2x}{2}\right)+c=\dfrac{x}{2}-\dfrac{\sin2x}{4}+c

\dfrac{x}{2}-\dfrac{\sin 2x}{4}+c

Check point

Evaluate \int\cos^{2}x\,dx.

Replace \cos^{2}x using the identity and split the integral.
\int\cos^{2}x\,dx=\int\dfrac{1+\cos2x}{2}\,dx=\dfrac12\int dx+\dfrac12\int\cos2x\,dx
Integrate each part.
=\dfrac{x}{2}+\dfrac{\sin2x}{4}+c

\dfrac{x}{2}+\dfrac{\sin 2x}{4}+c

Example 7 (ii)

Integrate \cos^{2}7x with respect to x.

Use the identity with the angle 7x, so the double angle is 14x.
\int\cos^{2}7x\,dx=\int\dfrac{1+\cos14x}{2}\,dx=\dfrac12\int dx+\dfrac12\int\cos14x\,dx
Integrate: \int\cos14x\,dx=\dfrac{\sin14x}{14}.
=\dfrac{x}{2}+\dfrac12\left(\dfrac{\sin14x}{14}\right)+c=\dfrac{x}{2}+\dfrac{\sin14x}{28}+c

\dfrac{x}{2}+\dfrac{\sin 14x}{28}+c

\int\sin^{2}kx\,dx=\dfrac{x}{2}-\dfrac{\sin2kx}{4k}+c,\qquad \int\cos^{2}kx\,dx=\dfrac{x}{2}+\dfrac{\sin2kx}{4k}+c

General forms for any constant k\neq0. In an exam, always show the identity step.

Other squares and products

The same idea works for other integrands: change them with an identity until every term is in the table.

Integrand	Identity	Integral
\tan^{2}x	\tan^{2}x=\sec^{2}x-1	\tan x-x+c
\cot^{2}x	\cot^{2}x=\operatorname{cosec}^{2}x-1	-\cot x-x+c
1+\tan^{2}x	1+\tan^{2}x=\sec^{2}x	\tan x+c
\sin x\cos x	\sin x\cos x=\tfrac12\sin2x	-\tfrac14\cos2x+c

3.3Integration by substitutionOften examined

Many integrals do not match any formula as they stand. Integration by substitution changes the variable so that the integral becomes a standard one. It is the chain rule used backwards.

The method works best when the integrand contains a function g(x) and its derivative g'(x) as a factor. Putting t=g(x) then turns the integrand into a simple function of t. There is no fixed rule for choosing the substitution; the right choice comes with practice.

Derivation Why substitution works

Aim. Show that \int f\big(g(x)\big)\,g'(x)\,dx=\int f(t)\,dt, where t=g(x).

Let F be an antiderivative of f, so F'(t)=f(t).
\int f(t)\,dt=F(t)+c
Differentiate F\big(g(x)\big) with the chain rule.
\dfrac{d}{dx}F\big(g(x)\big)=F'\big(g(x)\big)\,g'(x)=f\big(g(x)\big)\,g'(x)
Read this backwards: F\big(g(x)\big) is an antiderivative of f\big(g(x)\big)g'(x).
\int f\big(g(x)\big)\,g'(x)\,dx=F\big(g(x)\big)+c=F(t)+c=\int f(t)\,dt
In practice we write t=g(x) and dt=g'(x)\,dx, and replace both in the integral.
t=g(x),\qquad dt=g'(x)\,dx

Result\int f\big(g(x)\big)\,g'(x)\,dx=\int f(t)\,dt,\qquad t=g(x)

When the derivative of the inner function is present, the substitution removes both the inner function and its derivative, leaving an integral in t alone.

Choose t: usually the inner function (inside a bracket, a root, an exponent or a trigonometric function).
Differentiate: write dt=g'(x)\,dx, and solve for the part of the integrand you need, for example x\,dx=\tfrac12dt.
Replace every x and dx, so that only t and dt remain.
Integrate with respect to t.
Substitute back t=g(x) and add c.

Example 8

Evaluate \int3x^{2}\cos(x^{3})\,dx.

The inner function is x^{3} and its derivative 3x^{2} is present. Put t=x^{3}.
x^{3}=t\qquad(\text{i})
Differentiate (i).
3x^{2}\,dx=dt\qquad(\text{ii})
Substitute (i) and (ii) in the integral and integrate.
\int\cos(x^{3})\,3x^{2}\,dx=\int\cos t\,dt=\sin t+c
Substitute back t=x^{3}.
\int3x^{2}\cos(x^{3})\,dx=\sin(x^{3})+c

\sin(x^{3})+c

Example 9

Integrate \displaystyle\int\frac{e^{\tan^{-1}x}}{1+x^{2}}\,dx.

Let u=\tan^{-1}x. Its derivative is in the integrand.
u=\tan^{-1}x\;\Longrightarrow\;du=\dfrac{1}{1+x^{2}}\,dx
Substitute, integrate and substitute back.
\int\dfrac{e^{\tan^{-1}x}}{1+x^{2}}\,dx=\int e^{u}\,du=e^{u}+c=e^{\tan^{-1}x}+c

e^{\tan^{-1}x}+c

Check point

Integrate x\sin(x^{2}-3) with respect to x.

Put t=x^{2}-3. Then dt=2x\,dx, so x\,dx=\tfrac12dt.
t=x^{2}-3,\qquad x\,dx=\dfrac12\,dt
Substitute and integrate.
\int\sin(x^{2}-3)\,x\,dx=\dfrac12\int\sin t\,dt=-\dfrac12\cos t+c
Substitute back.
=-\dfrac12\cos(x^{2}-3)+c

-\dfrac{1}{2}\cos(x^{2}-3)+c

Example 12

Evaluate \int x(x^{2}-a^{2})^{3/2}\,dx.

Put u=x^{2}-a^{2}, the bracket. Then du=2x\,dx, so x\,dx=\tfrac12du.
x^{2}-a^{2}=u\;\Longrightarrow\;2x\,dx=du\;\Longrightarrow\;x\,dx=\dfrac{du}{2}
Substitute and use the power rule.
\int(x^{2}-a^{2})^{3/2}\,x\,dx=\dfrac12\int u^{3/2}\,du=\dfrac12\cdot\dfrac{u^{5/2}}{5/2}+c=\dfrac15u^{5/2}+c
Substitute back u=x^{2}-a^{2}.
=\dfrac15(x^{2}-a^{2})^{5/2}+c

\dfrac{1}{5}(x^{2}-a^{2})^{5/2}+c

Common substitutions. A good choice is one whose derivative is (a constant multiple of) another factor of the integrand.
Integrand contains	Try	Example
a bracket raised to a power	t= the bracket	\int x(x^{2}+1)^{7}dx: t=x^{2}+1
e^{g(x)}	t=g(x), the power	\int xe^{x^{2}}dx: t=x^{2}
a trigonometric function of g(x)	t=g(x)	\int x^{2}\cos(x^{3})\,dx: t=x^{3}
\sqrt{g(x)}	t=g(x)	\int x\sqrt{x^{2}+4}\,dx: t=x^{2}+4
\dfrac{f'(x)}{f(x)}	t=f(x)	\int\dfrac{2x}{x^{2}+3}\,dx=\ln(x^{2}+3)+c

3.3Trigonometric substitution and standard resultsOften examined

When an integrand contains \sqrt{a^{2}-x^{2}}, \sqrt{x^{2}-a^{2}}, \sqrt{x^{2}+a^{2}} or a^{2}+x^{2} (with a>0), a trigonometric substitution removes the square root. A Pythagorean identity turns the expression under the root into a perfect square.

Expression	Substitution	Identity used	Result
\sqrt{a^{2}-x^{2}}	x=a\sin\theta	1-\sin^{2}\theta=\cos^{2}\theta	\sqrt{a^{2}-x^{2}}=a\cos\theta
\sqrt{x^{2}-a^{2}}	x=a\sec\theta	\sec^{2}\theta-1=\tan^{2}\theta	\sqrt{x^{2}-a^{2}}=a\tan\theta
\sqrt{x^{2}+a^{2}} or a^{2}+x^{2}	x=a\tan\theta	1+\tan^{2}\theta=\sec^{2}\theta	\sqrt{x^{2}+a^{2}}=a\sec\theta

Reference triangles. Draw the triangle that matches the substitution, then read off the other trigonometric ratios in terms of x to return to the variable x.

Derivation Formula 3.5

Aim. Prove that \displaystyle\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\frac{x}{a}\right)+c using x=a\sin\theta, -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}.

Substitute x=a\sin\theta and differentiate to get dx.
x=a\sin\theta\;\Longrightarrow\;dx=a\cos\theta\,d\theta
Replace x and dx in the integral.
\int\dfrac{dx}{\sqrt{a^{2}-x^{2}}}=\int\dfrac{a\cos\theta\,d\theta}{\sqrt{a^{2}-a^{2}\sin^{2}\theta}}=\int\dfrac{a\cos\theta\,d\theta}{a\sqrt{1-\sin^{2}\theta}}
Use 1-\sin^{2}\theta=\cos^{2}\theta. For -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}, \cos\theta\ge0, so \sqrt{\cos^{2}\theta}=\cos\theta.
=\int\dfrac{a\cos\theta}{a\cos\theta}\,d\theta=\int d\theta=\theta+c
Return to x: \sin\theta=\dfrac{x}{a}, so \theta=\sin^{-1}\dfrac{x}{a}.
\int\dfrac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\dfrac{x}{a}\right)+c

Result\int\dfrac{1}{\sqrt{a^{2}-x^{2}}}\,dx=\sin^{-1}\left(\dfrac{x}{a}\right)+c

With a=1 this is the derivative of \sin^{-1}x read backwards. Once proved, the formula can be applied directly.

Derivation Formula 3.6

Aim. Prove that \displaystyle\int\sqrt{a^{2}-x^{2}}\,dx=\frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right)+\frac{x\sqrt{a^{2}-x^{2}}}{2}+c.

Substitute x=a\sin\theta, dx=a\cos\theta\,d\theta. As before, the root becomes a\cos\theta.
\int\sqrt{a^{2}-x^{2}}\,dx=\int a\cos\theta\cdot a\cos\theta\,d\theta=a^{2}\int\cos^{2}\theta\,d\theta
Lower the power with \cos^{2}\theta=\dfrac{1+\cos2\theta}{2} and integrate.
=\dfrac{a^{2}}{2}\int(1+\cos2\theta)\,d\theta=\dfrac{a^{2}}{2}\left(\theta+\dfrac{\sin2\theta}{2}\right)+c
Use \sin2\theta=2\sin\theta\cos\theta, so that everything is in \sin\theta and \cos\theta.
=\dfrac{a^{2}}{2}\theta+\dfrac{a^{2}}{2}\sin\theta\cos\theta+c
Return to x: \sin\theta=\dfrac{x}{a}, \cos\theta=\sqrt{1-\sin^{2}\theta}=\dfrac{\sqrt{a^{2}-x^{2}}}{a} and \theta=\sin^{-1}\dfrac{x}{a}.
=\dfrac{a^{2}}{2}\sin^{-1}\left(\dfrac{x}{a}\right)+\dfrac{a^{2}}{2}\cdot\dfrac{x}{a}\cdot\dfrac{\sqrt{a^{2}-x^{2}}}{a}+c=\dfrac{a^{2}}{2}\sin^{-1}\left(\dfrac{x}{a}\right)+\dfrac{x\sqrt{a^{2}-x^{2}}}{2}+c

Result\int\sqrt{a^{2}-x^{2}}\,dx=\dfrac{a^{2}}{2}\sin^{-1}\left(\dfrac{x}{a}\right)+\dfrac{x\sqrt{a^{2}-x^{2}}}{2}+c

This integral measures areas of parts of a circle of radius a. Section 3.4 proves the same result by integration by parts (Example 15).

Completing the square

If the expression under the root or in the denominator is a quadratic px^{2}+qx+r, complete the square first. It then takes the form a^{2}-u^{2} or u^{2}+a^{2} with u=x+k, and a standard formula applies with u in place of x (note du=dx).

Example 10

Evaluate \displaystyle\int\frac{1}{\sqrt{5-4x-x^{2}}}\,dx.

Complete the square: 5-4x-x^{2}=5+4-(x^{2}+4x+4)=9-(x+2)^{2}.
\int\dfrac{dx}{\sqrt{5-4x-x^{2}}}=\int\dfrac{dx}{\sqrt{9-(x^{2}+4x+4)}}=\int\dfrac{dx}{\sqrt{3^{2}-(x+2)^{2}}}
This is Formula 3.5 with a=3 and x+2 in place of x.
=\sin^{-1}\left(\dfrac{x+2}{3}\right)+c

\sin^{-1}\left(\dfrac{x+2}{3}\right)+c

The same answer comes from the substitution x+2=3\sin\theta.

Derivation Formula 3.7

Aim. Prove that \displaystyle\int\frac{dx}{\sqrt{x^{2}-a^{2}}}=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+C using x=a\sec\theta.

Substitute x=a\sec\theta and differentiate.
x=a\sec\theta\;\Longrightarrow\;dx=a\sec\theta\tan\theta\,d\theta
Simplify the root with \sec^{2}\theta-1=\tan^{2}\theta (take 0\le\theta<\tfrac{\pi}{2}, so \tan\theta\ge0).
\sqrt{x^{2}-a^{2}}=\sqrt{a^{2}\sec^{2}\theta-a^{2}}=\sqrt{a^{2}(\sec^{2}\theta-1)}=a\tan\theta
Replace in the integral and cancel a\tan\theta.
\int\dfrac{a\sec\theta\tan\theta}{a\tan\theta}\,d\theta=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+c
Return to x: \sec\theta=\dfrac{x}{a} and \tan\theta=\sqrt{\sec^{2}\theta-1}=\dfrac{\sqrt{x^{2}-a^{2}}}{a}.
=\ln\left|\dfrac{x}{a}+\dfrac{\sqrt{x^{2}-a^{2}}}{a}\right|+c=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|-\ln a+c
c-\ln a is just another constant; call it C.
=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+C

Result\int\dfrac{1}{\sqrt{x^{2}-a^{2}}}\,dx=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+C

The constant -\ln a is absorbed into C. The book notes that the substitution x=a\cosh\theta also works; it gives the equivalent form \cosh^{-1}\left(\tfrac{x}{a}\right)+C.

Derivation Formula 3.8

Aim. Prove that \displaystyle\int\frac{dx}{\sqrt{x^{2}+a^{2}}}=\ln\left(x+\sqrt{x^{2}+a^{2}}\right)+C using x=a\tan\theta.

Substitute x=a\tan\theta with -\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2}, so that \sec\theta>0.
x=a\tan\theta\;\Longrightarrow\;dx=a\sec^{2}\theta\,d\theta
Simplify the root with 1+\tan^{2}\theta=\sec^{2}\theta.
\sqrt{x^{2}+a^{2}}=\sqrt{a^{2}\tan^{2}\theta+a^{2}}=a\sec\theta
Replace and cancel.
\int\dfrac{a\sec^{2}\theta}{a\sec\theta}\,d\theta=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+c
Return to x: \tan\theta=\dfrac{x}{a} and \sec\theta=\sqrt{1+\tan^{2}\theta}=\dfrac{\sqrt{a^{2}+x^{2}}}{a}.
=\ln\left|\dfrac{x+\sqrt{a^{2}+x^{2}}}{a}\right|+c=\ln\left(x+\sqrt{a^{2}+x^{2}}\right)-\ln a+c
Absorb -\ln a into the constant. No absolute value is needed, because x+\sqrt{x^{2}+a^{2}}>0 for every x.
=\ln\left(x+\sqrt{x^{2}+a^{2}}\right)+C

Result\int\dfrac{1}{\sqrt{x^{2}+a^{2}}}\,dx=\ln\left(x+\sqrt{x^{2}+a^{2}}\right)+C

The 2026 old-syllabus paper (Curriculum 2000, Q.2 (iii)) asked exactly this with a=5.

Derivation Formula 3.9

Aim. Prove that \displaystyle\int\frac{dx}{a^{2}+x^{2}}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+c using x=a\tan\theta.

Substitute x=a\tan\theta, dx=a\sec^{2}\theta\,d\theta, and simplify the denominator.
a^{2}+x^{2}=a^{2}+a^{2}\tan^{2}\theta=a^{2}\sec^{2}\theta
Replace and cancel \sec^{2}\theta.
\int\dfrac{a\sec^{2}\theta\,d\theta}{a^{2}\sec^{2}\theta}=\dfrac1a\int d\theta=\dfrac{\theta}{a}+c
Return to x: \theta=\tan^{-1}\dfrac{x}{a}.
=\dfrac1a\tan^{-1}\left(\dfrac{x}{a}\right)+c

Result\int\dfrac{1}{a^{2}+x^{2}}\,dx=\dfrac1a\tan^{-1}\left(\dfrac{x}{a}\right)+c

The book only says this 'can easily be proved by substituting x=a\tan\theta'; the steps above are that proof. Remember the factor \tfrac1a in front.

Example 11

Evaluate \displaystyle\int\frac{1}{x^{2}+4x+5}\,dx.

Complete the square in the denominator.
x^{2}+4x+5=(x^{2}+4x+4)+1=(x+2)^{2}+1^{2}
Apply Formula 3.9 with a=1 and x+2 in place of x.
\int\dfrac{dx}{(x+2)^{2}+1^{2}}=\dfrac11\tan^{-1}\left(\dfrac{x+2}{1}\right)+c=\tan^{-1}(x+2)+c

\tan^{-1}(x+2)+c

The book notes that the substitution x+2=\tan\theta gives the same result.

3.4Integration by partsOften examined

Integration by parts integrates a product of two functions of different types, such as xe^{x}, x^{2}\ln x or x\tan^{-1}x, when substitution does not help. It reduces the integral to a standard form. It is the product rule for derivatives used backwards.

Derivation The integration by parts formula

Aim. Derive \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int\left[f'(x)\int g(x)\,dx\right]dx.

Let G(x)=\int g(x)\,dx, so that G'(x)=g(x). Differentiate the product f(x)G(x) with the product rule.
\dfrac{d}{dx}\big[f(x)G(x)\big]=f(x)G'(x)+f'(x)G(x)=f(x)g(x)+f'(x)G(x)
Integrate both sides with respect to x. The left side is the integral of a derivative, so it gives back f(x)G(x).
f(x)G(x)=\int f(x)g(x)\,dx+\int f'(x)G(x)\,dx
Make the integral of the product the subject.
\int f(x)g(x)\,dx=f(x)G(x)-\int f'(x)G(x)\,dx
Put back G(x)=\int g(x)\,dx.
\int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int\left[f'(x)\int g(x)\,dx\right]dx\qquad(1)

Result\int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int\left[f'(x)\int g(x)\,dx\right]dx

We differentiate the first function f and integrate the second function g. It helps when f' is simpler than f, so that the new integral is easier than the original. In u,v notation the same formula reads \int u\,dv=uv-\int v\,du.

In words: first function \times integral of the second, minus the integral of (derivative of the first \times integral of the second).

Choosing the first function

Take as second function g the one you can integrate easily, and as first function f the one whose derivative is simpler.
If one function is a power of x or a polynomial, it is usually the first function. But if the other function is an inverse trigonometric function or a logarithm, that one becomes the first function (we cannot easily integrate \ln x or \tan^{-1}x, but we can differentiate them).
If the product is an exponential times a trigonometric function, either can be the first function.
Do not add a constant when you integrate the second function inside the formula; one constant is added at the very end.
Integration by parts does not work for every product. For example, it cannot be used for \int\sqrt{x}\sin x\,dx.

Example 13

Evaluate \int xe^{x}\,dx.

Take x as the first function (its derivative, 1, is simpler) and e^{x} as the second function.
\int xe^{x}\,dx=x\int e^{x}dx-\int\left[\dfrac{d}{dx}(x)\int e^{x}dx\right]dx
Simplify with \int e^{x}dx=e^{x} and \dfrac{d}{dx}(x)=1.
=xe^{x}-\int1\cdot e^{x}\,dx=xe^{x}-e^{x}+c

xe^{x}-e^{x}+c

Example 14 (i)

Evaluate \int x^{2}\ln x\,dx.

Take \ln x as the first function (easy to differentiate, hard to integrate) and x^{2} as the second.
\int x^{2}\ln x\,dx=\ln x\int x^{2}dx-\int\left[\dfrac{d}{dx}(\ln x)\int x^{2}dx\right]dx
Use \int x^{2}dx=\dfrac{x^{3}}{3} and \dfrac{d}{dx}\ln x=\dfrac1x.
=\ln x\cdot\dfrac{x^{3}}{3}-\int\dfrac1x\cdot\dfrac{x^{3}}{3}\,dx=\dfrac{x^{3}\ln x}{3}-\dfrac13\int x^{2}\,dx
Integrate the remaining power.
=\dfrac{x^{3}\ln x}{3}-\dfrac13\cdot\dfrac{x^{3}}{3}+c=\dfrac{x^{3}\ln x}{3}-\dfrac{x^{3}}{9}+c

\dfrac{x^{3}\ln x}{3}-\dfrac{x^{3}}{9}+c

Example 14 (ii)

Evaluate \int x\tan^{-1}x\,dx.

Take \tan^{-1}x as the first function and x as the second.
\int x\tan^{-1}x\,dx=\tan^{-1}x\int x\,dx-\int\left[\dfrac{d}{dx}(\tan^{-1}x)\int x\,dx\right]dx
Use \int x\,dx=\dfrac{x^{2}}{2} and \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}}.
=\dfrac{x^{2}\tan^{-1}x}{2}-\dfrac12\int\dfrac{x^{2}}{x^{2}+1}\,dx
Divide: \dfrac{x^{2}}{x^{2}+1}=\dfrac{(x^{2}+1)-1}{x^{2}+1}=1-\dfrac{1}{x^{2}+1}.
=\dfrac{x^{2}\tan^{-1}x}{2}-\dfrac12\int\left(1-\dfrac{1}{x^{2}+1}\right)dx=\dfrac{x^{2}\tan^{-1}x}{2}-\dfrac12\left(x-\tan^{-1}x\right)+c

\dfrac{x^{2}\tan^{-1}x}{2}-\dfrac{x}{2}+\dfrac{1}{2}\tan^{-1}x+c

A single function: the integral of ln x

Evaluate \int\ln x\,dx by parts.

Write \ln x=(\ln x)(1). Take \ln x as the first function and 1 as the second.
\int\ln x\cdot1\,dx=\ln x\int1\,dx-\int\left[\dfrac1x\int1\,dx\right]dx=x\ln x-\int\dfrac1x\cdot x\,dx
The remaining integral is \int1\,dx=x.
=x\ln x-x+c

x\ln x-x+c

The same 'times 1' trick integrates \sin^{-1}x, \cos^{-1}x and \tan^{-1}x.

Integrals that come back (Examples 15 and 16)

Sometimes integration by parts brings back the integral we started with. Call it I, collect the I terms on one side and solve for I.

Example 15 (i)

Apply integration by parts to evaluate \int\sqrt{a^{2}-x^{2}}\,dx.

Write the integrand as \sqrt{a^{2}-x^{2}}\cdot1. Take \sqrt{a^{2}-x^{2}} as the first function and 1 as the second.
I=\int\sqrt{a^{2}-x^{2}}\,dx=\sqrt{a^{2}-x^{2}}\int1\,dx-\int\left[\dfrac{d}{dx}\sqrt{a^{2}-x^{2}}\int1\,dx\right]dx
Use \dfrac{d}{dx}\sqrt{a^{2}-x^{2}}=\dfrac{-2x}{2\sqrt{a^{2}-x^{2}}}=\dfrac{-x}{\sqrt{a^{2}-x^{2}}}.
I=x\sqrt{a^{2}-x^{2}}-\int\dfrac{-x}{\sqrt{a^{2}-x^{2}}}\cdot x\,dx=x\sqrt{a^{2}-x^{2}}-\int\dfrac{-x^{2}}{\sqrt{a^{2}-x^{2}}}\,dx
Write -x^{2}=(a^{2}-x^{2})-a^{2} and split the fraction.
I=x\sqrt{a^{2}-x^{2}}-\int\dfrac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}\,dx+\int\dfrac{a^{2}}{\sqrt{a^{2}-x^{2}}}\,dx
The first new integral is \int\sqrt{a^{2}-x^{2}}\,dx=I again. The second is Formula 3.5.
I=x\sqrt{a^{2}-x^{2}}-I+a^{2}\sin^{-1}\left(\dfrac{x}{a}\right)+c
Collect the I terms and divide by 2. Half of an arbitrary constant is another arbitrary constant, C.
2I=x\sqrt{a^{2}-x^{2}}+a^{2}\sin^{-1}\left(\dfrac{x}{a}\right)+c\;\Longrightarrow\;I=\dfrac{x\sqrt{a^{2}-x^{2}}}{2}+\dfrac{a^{2}}{2}\sin^{-1}\left(\dfrac{x}{a}\right)+C

\dfrac{x\sqrt{a^{2}-x^{2}}}{2}+\dfrac{a^{2}}{2}\sin^{-1}\left(\dfrac{x}{a}\right)+C

Example 15 (ii)

Apply integration by parts to evaluate \int\sqrt{a^{2}+x^{2}}\,dx.

Take \sqrt{a^{2}+x^{2}} as the first function and 1 as the second. The derivative of the first is \dfrac{x}{\sqrt{a^{2}+x^{2}}}.
I=\int\sqrt{a^{2}+x^{2}}\,dx=x\sqrt{a^{2}+x^{2}}-\int\dfrac{x^{2}}{\sqrt{a^{2}+x^{2}}}\,dx
Write x^{2}=(a^{2}+x^{2})-a^{2} and split.
I=x\sqrt{a^{2}+x^{2}}-\int\sqrt{a^{2}+x^{2}}\,dx+\int\dfrac{a^{2}}{\sqrt{a^{2}+x^{2}}}\,dx
The first integral is I. The second is Formula 3.8: \int\dfrac{dx}{\sqrt{x^{2}+a^{2}}}=\ln\left(x+\sqrt{x^{2}+a^{2}}\right).
I=x\sqrt{a^{2}+x^{2}}-I+a^{2}\ln\left(x+\sqrt{a^{2}+x^{2}}\right)+c
Collect the I terms and divide by 2.
I=\dfrac{x\sqrt{a^{2}+x^{2}}}{2}+\dfrac{a^{2}}{2}\ln\left(x+\sqrt{a^{2}+x^{2}}\right)+C

\dfrac{x\sqrt{a^{2}+x^{2}}}{2}+\dfrac{a^{2}}{2}\ln\left(x+\sqrt{a^{2}+x^{2}}\right)+C

Check point

Using integration by parts, prove that \int\sqrt{x^{2}-a^{2}}\,dx=\dfrac{x\sqrt{x^{2}-a^{2}}}{2}-\dfrac{a^{2}}{2}\cosh^{-1}\left(\dfrac{x}{a}\right)+c.

Take \sqrt{x^{2}-a^{2}} as the first function and 1 as the second. The derivative of the first is \dfrac{x}{\sqrt{x^{2}-a^{2}}}.
I=x\sqrt{x^{2}-a^{2}}-\int\dfrac{x^{2}}{\sqrt{x^{2}-a^{2}}}\,dx
Write x^{2}=(x^{2}-a^{2})+a^{2} and split.
I=x\sqrt{x^{2}-a^{2}}-\int\sqrt{x^{2}-a^{2}}\,dx-a^{2}\int\dfrac{dx}{\sqrt{x^{2}-a^{2}}}
The first integral is I; the second is Formula 3.7.
2I=x\sqrt{x^{2}-a^{2}}-a^{2}\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+c
Divide by 2. Since \cosh^{-1}\dfrac{x}{a}=\ln\dfrac{x+\sqrt{x^{2}-a^{2}}}{a}, the logarithm differs from \cosh^{-1}\dfrac{x}{a} only by the constant \ln a, which joins the constant.
I=\dfrac{x\sqrt{x^{2}-a^{2}}}{2}-\dfrac{a^{2}}{2}\cosh^{-1}\left(\dfrac{x}{a}\right)+C

Example 16

Apply integration by parts to evaluate \int e^{ax}\sin bx\,dx.

Call the integral I. Take \sin bx as the first function and e^{ax} as the second, with \int e^{ax}dx=\dfrac{e^{ax}}{a}.
I=\sin bx\left(\dfrac{e^{ax}}{a}\right)-\int(b\cos bx)\left(\dfrac{e^{ax}}{a}\right)dx=\dfrac{e^{ax}\sin bx}{a}-\dfrac{b}{a}\int e^{ax}\cos bx\,dx
Apply parts again to the new integral, now with \cos bx first and e^{ax} second.
\int e^{ax}\cos bx\,dx=\dfrac{e^{ax}\cos bx}{a}-\int(-b\sin bx)\dfrac{e^{ax}}{a}\,dx=\dfrac{e^{ax}\cos bx}{a}+\dfrac{b}{a}I
Substitute back. The original integral I appears on the right.
I=\dfrac{e^{ax}\sin bx}{a}-\dfrac{b}{a^{2}}e^{ax}\cos bx-\dfrac{b^{2}}{a^{2}}I+c
Collect the I terms: I+\dfrac{b^{2}}{a^{2}}I=\dfrac{a^{2}+b^{2}}{a^{2}}I.
\left(\dfrac{a^{2}+b^{2}}{a^{2}}\right)I=e^{ax}\left[\dfrac1a\sin bx-\dfrac{b}{a^{2}}\cos bx\right]+c
Multiply both sides by \dfrac{a^{2}}{a^{2}+b^{2}}.
I=\dfrac{e^{ax}}{a^{2}+b^{2}}\big(a\sin bx-b\cos bx\big)+C

\dfrac{e^{ax}}{a^{2}+b^{2}}(a\sin bx-b\cos bx)+C

\int e^{ax}\cos bx\,dx=\dfrac{e^{ax}}{a^{2}+b^{2}}\big(a\cos bx+b\sin bx\big)+C

The companion result, found in the same way.

Integration by parts with limits

For a definite integral, apply the limits to the product term as well as to the remaining integral:

\int_{a}^{b}f(x)g(x)\,dx=\left[f(x)\int g(x)\,dx\right]_{a}^{b}-\int_{a}^{b}\left[f'(x)\int g(x)\,dx\right]dx

By parts with limits

Evaluate \int_{0}^{1}xe^{x}\,dx.

From Example 13, an antiderivative of xe^{x} is xe^{x}-e^{x}.
\int_{0}^{1}xe^{x}\,dx=\Big[xe^{x}-e^{x}\Big]_{0}^{1}
Upper limit minus lower limit.
=(1\cdot e-e)-(0-e^{0})=0-(-1)=1

3.5Integration by partial fractionsOften examined

Adding \dfrac{3}{x+4}+\dfrac{4}{x+2} over a common denominator gives the single fraction \dfrac{7x+22}{(x+4)(x+2)}. Partial fractions reverse this: they split a rational function into simple fractions. Each simple fraction integrates to a logarithm, a power, or an inverse tangent.

Introductory example

Evaluate \displaystyle\int\frac{7x+22}{(x+4)(x+2)}\,dx.

Use the split shown above.
\dfrac{7x+22}{(x+4)(x+2)}=\dfrac{3}{x+4}+\dfrac{4}{x+2}
Integrate each fraction with \int\dfrac{dx}{x+k}=\ln|x+k|.
\int\dfrac{7x+22}{(x+4)(x+2)}\,dx=3\int\dfrac{dx}{x+4}+4\int\dfrac{dx}{x+2}=3\ln|x+4|+4\ln|x+2|+c

3\ln|x+4|+4\ln|x+2|+c

Proper rational fraction: \dfrac{P(x)}{Q(x)} with the degree of P less than the degree of Q. Only a proper fraction can be split directly.
Improper rational fraction: The degree of P is greater than or equal to the degree of Q. Divide first to get a polynomial plus a proper fraction.

The form of the partial fractions depends only on the factors of the denominator.
Factor in the denominator	Partial fractions to use
distinct linear factors (x-a)(x-b)	\dfrac{A}{x-a}+\dfrac{B}{x-b}
repeated linear factor (x-a)^{3}	\dfrac{A}{x-a}+\dfrac{B}{(x-a)^{2}}+\dfrac{C}{(x-a)^{3}}
quadratic factor x^{2}+px+q with no real roots (p^{2}<4q)	\dfrac{Bx+C}{x^{2}+px+q}

If the fraction is improper, divide first.
Factorise the denominator completely.
Write the partial fractions with unknown constants A, B, C, and so on.
Multiply through by the denominator. Find the constants by putting x equal to the roots of the linear factors, and by comparing coefficients of like powers of x.
Integrate each term: a linear factor gives a logarithm, a repeated factor gives a power, a quadratic factor gives a logarithm plus an inverse tangent.

Example 17 (improper fraction)

Evaluate \displaystyle\int\frac{x^{3}-2x}{x^{2}+3x+2}\,dx.

The numerator has degree 3 and the denominator degree 2, so divide first. Long division gives quotient x-3 and remainder 5x+6.
\dfrac{x^{3}-2x}{x^{2}+3x+2}=x-3+\dfrac{5x+6}{x^{2}+3x+2}\qquad(\text{i})
Factorise the denominator and write the partial fractions.
\dfrac{5x+6}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}
Multiply by (x+1)(x+2) to clear the fractions.
5x+6=A(x+2)+B(x+1)\qquad(\text{ii})
Put x=-2 in (ii): -4=-B, so B=4. Put x=-1: 1=A.
A=1,\qquad B=4
Integrate every term.
\int\left[x-3+\dfrac{1}{x+1}+\dfrac{4}{x+2}\right]dx=\dfrac{x^{2}}{2}-3x+\ln|x+1|+4\ln|x+2|+c

\dfrac{x^{2}}{2}-3x+\ln|x+1|+4\ln|x+2|+c

Example 18 (repeated linear factor)

Evaluate \displaystyle\int\frac{x^{2}+2x+4}{(x+1)^{3}}\,dx.

The factor (x+1) is repeated three times, so use one fraction for each power.
\dfrac{x^{2}+2x+4}{(x+1)^{3}}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^{2}}+\dfrac{C}{(x+1)^{3}}
Multiply by (x+1)^{3} and expand.
x^{2}+2x+4=A(x+1)^{2}+B(x+1)+C=Ax^{2}+(2A+B)x+(A+B+C)
Compare coefficients of x^{2}, of x, and the constant terms.
A=1,\quad 2A+B=2,\quad A+B+C=4\;\Longrightarrow\;A=1,\ B=0,\ C=3
Integrate. The last term is a power: \int(x+1)^{-3}dx=\dfrac{(x+1)^{-2}}{-2}.
\int\left[\dfrac{1}{x+1}+\dfrac{3}{(x+1)^{3}}\right]dx=\ln|x+1|-\dfrac32(x+1)^{-2}+c=\ln|x+1|-\dfrac{3}{2(x+1)^{2}}+c

\ln|x+1|-\dfrac{3}{2(x+1)^{2}}+c

Example 19 (quadratic factor)

Evaluate \displaystyle\int\frac{3x^{2}+5x+3}{(x+2)(x^{2}+1)}\,dx.

x^{2}+1 has no real factors, so its partial fraction has a linear numerator.
\dfrac{3x^{2}+5x+3}{(x+2)(x^{2}+1)}=\dfrac{A}{x+2}+\dfrac{Bx+C}{x^{2}+1}
Multiply by (x+2)(x^{2}+1) and collect powers of x.
3x^{2}+5x+3=A(x^{2}+1)+(Bx+C)(x+2)=(A+B)x^{2}+(2B+C)x+(A+2C)
Compare coefficients.
A+B=3,\qquad 2B+C=5,\qquad A+2C=3
Solve. The first and third give B=2C; then the second gives 5C=5.
C=1,\quad B=2,\quad A=1
Integrate. Split \dfrac{2x+1}{x^{2}+1}=\dfrac{2x}{x^{2}+1}+\dfrac{1}{x^{2}+1}: the first part is 'derivative over function', the second gives \tan^{-1}x.
\int\left(\dfrac{1}{x+2}+\dfrac{2x}{x^{2}+1}+\dfrac{1}{x^{2}+1}\right)dx=\ln|x+2|+\ln(x^{2}+1)+\tan^{-1}x+c

\ln|x+2|+\ln(x^{2}+1)+\tan^{-1}x+c

Check point

Evaluate \displaystyle\int\frac{2x+1}{(x-1)(x+3)}\,dx.

Write the partial fractions and clear the denominator.
\dfrac{2x+1}{(x-1)(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{x+3}\;\Longrightarrow\;2x+1=A(x+3)+B(x-1)
Put x=1: 3=4A. Put x=-3: -5=-4B.
A=\dfrac34,\qquad B=\dfrac54
Integrate each term.
\int\left(\dfrac{3/4}{x-1}+\dfrac{5/4}{x+3}\right)dx=\dfrac34\ln|x-1|+\dfrac54\ln|x+3|+c

\dfrac{3}{4}\ln|x-1|+\dfrac{5}{4}\ln|x+3|+c

Integrating a quadratic-factor term

For \displaystyle\int\frac{Bx+C}{x^{2}+px+q}\,dx, write the numerator as a multiple of the derivative 2x+p plus a constant. The first part gives a logarithm; complete the square in the second part to get an inverse tangent.

\dfrac{Bx+C}{x^{2}+px+q}=\dfrac{B}{2}\cdot\dfrac{2x+p}{x^{2}+px+q}+\dfrac{C-\frac{Bp}{2}}{\left(x+\frac{p}{2}\right)^{2}+\left(q-\frac{p^{2}}{4}\right)}

\int\dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right|+c

A useful result from the partial fractions \dfrac{1}{x^{2}-a^{2}}=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right).

3.6The definite integral

The definite integral links integration with area, volume, length, density, probability and work. The idea: cut the region under a curve into thin rectangles, add their areas, and let the rectangles become infinitely thin.

Partition of an interval (3.6.1)

Partition: A set of points a=x_{0}<x_{1}<x_{2}<\dots<x_{n-1}<x_{n}=b that divides [a,b] into n subintervals of lengths \Delta x_{1}=x_{1}-x_{0}, \Delta x_{2}=x_{2}-x_{1}, and so on up to \Delta x_{n}=x_{n}-x_{n-1}.
Regular partition: A partition in which all subintervals have the same length, \Delta x=\dfrac{b-a}{n}. The widths tend to zero as n is made large.

Rectangles under a curve

A regular partition of [a,b] into n=8 strips. Each rectangle has width \Delta x and height f(x_{k}), the value at its right end. The sum of their areas approximates the area under the curve, and the error shrinks as n grows.

Each strip is close to a rectangle. The first rectangle has width \Delta x_{1} and height f(x_{1}), so its area is f(x_{1})\Delta x_{1}. Adding all n rectangles:

\text{Area}\approx f(x_{1})\Delta x_{1}+f(x_{2})\Delta x_{2}+\dots+f(x_{n})\Delta x_{n}=\sum_{k=1}^{n}f(x_{k})\Delta x_{k}\qquad(\text{i})

This sum (a Riemann sum) only approximates the area. As the rectangles get thinner the error shrinks, and the exact area is the limit:

A=\lim_{\Delta x\to0}\sum_{k=1}^{n}f(x_{k})\Delta x_{k}\qquad(\text{ii})

Definite integral (Definition 3.2): A function f is integrable on the closed interval [a,b] if the limit \displaystyle\lim_{\Delta x\to0}\sum_{k=1}^{n}f(x_{k})\Delta x_{k} exists and does not depend on the choice of partition or of the points x_{k}. The limit is written \displaystyle\int_{a}^{b}f(x)\,dx and is called the definite integral of f from a to b.
Limits of integration: a is the lower limit and b is the upper limit of integration; f(x) is the integrand.
Value of the integral: The number that \int_{a}^{b}f(x)\,dx equals. A definite integral is a number, not a family of functions, so it has no +c.

\int_{a}^{b}f(x)\,dx=\lim_{\Delta x\to0}\sum_{k=1}^{n}f(x_{k})\Delta x_{k}\qquad(\text{iii})

Theorem 3.2. If f is continuous on [a,b], then f is integrable on [a,b], and \int_{a}^{b}f(x)\,dx is the net signed area between the curve and the x-axis: area above the axis counts as positive and area below the axis counts as negative.

	Indefinite integral	Definite integral
Written as	\int f(x)\,dx	\int_{a}^{b}f(x)\,dx
Result	a family of functions F(x)+c	a single number
Constant c	always added	not needed (it cancels)
Meaning	all antiderivatives of f	net signed area from a to b

In the simplest cases a definite integral can be worked out from the area of a familiar shape.

Example 20

Sketch the region whose area is represented by the definite integral, and evaluate the integral using an appropriate formula from geometry: (i) \int_{1}^{5}3\,dx (ii) \int_{-2}^{2}(x+3)\,dx (iii) \int_{0}^{1}\sqrt{1-x^{2}}\,dx.

(i)

The graph of the integrand is the horizontal line y=3. The region is a rectangle of height 3 over the interval from 1 to 5, so its width is 4.

\int_{1}^{5}3\,dx=\text{area of rectangle}=4\times3=12

(ii)

The graph is the line y=x+3. At x=-2, y=1; at x=2, y=5. The region is a trapezium with parallel sides 1 and 5 and width 4.

\int_{-2}^{2}(x+3)\,dx=\text{area of trapezium}=\dfrac12(1+5)(4)=12

(iii)

y=\sqrt{1-x^{2}} is the upper half of the circle x^{2}+y^{2}=1 (radius 1, centre at the origin). From x=0 to x=1 the region is a quarter circle.

\int_{0}^{1}\sqrt{1-x^{2}}\,dx=\text{area of quarter circle}=\dfrac14\pi(1)^{2}=\dfrac{\pi}{4}

Example 21

Evaluate (i) \int_{0}^{1}(x-1)\,dx (ii) \int_{0}^{2}(x-1)\,dx.

A_{1} lies below the x-axis and A_{2} above it. Each is a triangle with base 1 and height 1.

(i) On [0,1] the line y=x-1 runs from -1 up to 0, so the region is the triangle A_{1} below the axis. Its area is \tfrac12(1)(1)=\tfrac12, and the integral counts it as negative.
\int_{0}^{1}(x-1)\,dx=-\dfrac12
(ii) Split at x=1, where the line crosses the axis. A_{2} on [1,2] is above the axis with area \tfrac12.
\int_{0}^{2}(x-1)\,dx=\int_{0}^{1}(x-1)\,dx+\int_{1}^{2}(x-1)\,dx=-\dfrac12+\dfrac12=0
The integral (net signed area) is 0, but the total area of the two triangles is \tfrac12+\tfrac12=1 square unit.
\text{total area}=|A_{1}|+|A_{2}|=\dfrac12+\dfrac12=1

3.7Properties of the definite integralOften examined

For a<b the following facts hold. They follow from the area meaning of the integral, and also from the Fundamental Theorem (Section 3.8).

Property	Statement	Example
(i) Equal limits	\int_{a}^{a}f(x)\,dx=0	\int_{2}^{2}x\,dx=0
(ii) Reversing the limits	\int_{b}^{a}f(x)\,dx=-\int_{a}^{b}f(x)\,dx	\int_{1}^{0}(x-1)\,dx=-\int_{0}^{1}(x-1)\,dx=\tfrac12
(iii) Splitting at a point c	\int_{a}^{b}f(x)\,dx=\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx	\int_{0}^{2}(x-1)\,dx=\int_{0}^{1}(x-1)\,dx+\int_{1}^{2}(x-1)\,dx

Property (i) says there is no area above a single point. Property (ii) says that interchanging the limits reverses the sign of the integral. Property (iii) is used to split a region where the curve crosses the x-axis.

Theorem 3.3

If f and g are integrable on [a,b] and c is a constant, then cf, f+g and f-g are integrable on [a,b], and:

(i) \int_{a}^{b}c\,f(x)\,dx=c\int_{a}^{b}f(x)\,dx (a constant factor comes out; the limits are not affected).
(ii) \int_{a}^{b}[f(x)\pm g(x)]\,dx=\int_{a}^{b}f(x)\,dx\pm\int_{a}^{b}g(x)\,dx (the integral of a sum or difference is the sum or difference of the integrals).
(iii) \int_{a}^{b}[c\,f(x)\pm d\,g(x)]\,dx=c\int_{a}^{b}f(x)\,dx\pm d\int_{a}^{b}g(x)\,dx.

Example 22

Find (i) \int_{-1}^{4}[2f(x)+5g(x)]\,dx if \int_{-1}^{4}f(x)\,dx=2 and \int_{-1}^{4}g(x)\,dx=4; (ii) \int_{-1}^{3}4f(x)\,dx if \int_{-1}^{2}f(x)\,dx=3 and \int_{2}^{3}f(x)\,dx=1.

(i) Split with Theorem 3.3 and take out the constants.
\int_{-1}^{4}[2f(x)+5g(x)]\,dx=2\int_{-1}^{4}f(x)\,dx+5\int_{-1}^{4}g(x)\,dx=2(2)+5(4)=24
(ii) Take out 4, then split the interval at x=2 (property (iii)).
\int_{-1}^{3}4f(x)\,dx=4\left[\int_{-1}^{2}f(x)\,dx+\int_{2}^{3}f(x)\,dx\right]=4(3+1)=16

Two more useful properties

Dummy variable. The letter used for the variable does not change the value: \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(t)\,dt. For example \int_{1}^{2}x\,dx=\int_{1}^{2}t\,dt=\tfrac32.
Even and odd functions. If f(-x)=f(x) (even function), then \int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx. If f(-x)=-f(x) (odd function), then \int_{-a}^{a}f(x)\,dx=0.

y=x^{3} is odd. The region A_{1} below the axis on [-1.5,0] matches A_{2} above the axis on [0,1.5], so \int_{-1.5}^{1.5}x^{3}\,dx=0, although the total area is not zero.

3.8Fundamental theorem of calculusOften examined

Working out a limit of Riemann sums is slow. The Fundamental Theorem of Calculus gives a quick method: find any antiderivative, and subtract its values at the two limits. It joins the two halves of the unit: definite integrals (areas) and indefinite integrals (antiderivatives).

Derivation The Fundamental Theorem of Calculus

Aim. Show that \int_{a}^{b}f(x)\,dx=F(b)-F(a), where F is any antiderivative of f on [a,b].

The area function A(x) is the area under y=f(x) from a to x. Moving from x to x+h adds a thin strip whose area is about f(x)\,h (outlined), so A'(x)=f(x).

Let f be continuous and non-negative on [a,b]. The area under its graph from a to b is the definite integral. Let A(x) be the area from a up to a variable point x.
A=\int_{a}^{b}f(x)\,dx\qquad(\text{i})
Two values of A(x) are known at once. The area from a to a lies above a single point, so it is zero; the area from a to b is A.
A(a)=0,\qquad A(b)=A
When x increases by a small amount h, the area grows by a thin strip of width h and height about f(x). Divide by h and let h\to0.
A(x+h)-A(x)\approx f(x)\,h\;\Longrightarrow\;A'(x)=\lim_{h\to0}\dfrac{A(x+h)-A(x)}{h}=f(x)
So A(x) is an antiderivative of f. Any other antiderivative F differs from it by a constant.
F(x)=A(x)+c\qquad(\text{ii})
Put x=a and x=b in (ii).
F(a)=A(a)+c\quad(\text{iii}),\qquad F(b)=A(b)+c\quad(\text{iv})
Subtract (iii) from (iv). The constant c cancels.
F(b)-F(a)=\big[A(b)+c\big]-\big[A(a)+c\big]=A(b)-A(a)=A-0=A
Compare with (i).
\int_{a}^{b}f(x)\,dx=F(b)-F(a)\qquad(\text{v})

Result\int_{a}^{b}f(x)\,dx=\Big[F(x)\Big]_{x=a}^{x=b}=F(b)-F(a)

To evaluate a definite integral, find any antiderivative, then subtract its value at the lower limit from its value at the upper limit. The constant of integration always cancels, so it is left out.

Statement. If f is continuous on [a,b] and F is an antiderivative of f on [a,b], then \displaystyle\int_{a}^{b}f(x)\,dx=F(b)-F(a). Here a and b are values of the variable x.

\dfrac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x)

The same argument shows that differentiating an integral with respect to its upper limit gives back the integrand. For example \dfrac{d}{dx}\int_{-2}^{x}t^{3}\,dt=x^{3}.

Example 23

Evaluate \int_{1}^{3}x\,dx.

An antiderivative of x is \dfrac{x^{2}}{2}. Apply the upper limit, then the lower limit.
\int_{1}^{3}x\,dx=\left[\dfrac{x^{2}}{2}\right]_{1}^{3}=\dfrac{3^{2}}{2}-\dfrac{1^{2}}{2}=\dfrac92-\dfrac12=4

Example 24

Evaluate \int_{-2}^{2}(3x^{2}-x+1)\,dx.

Find an antiderivative term by term.
\int_{-2}^{2}(3x^{2}-x+1)\,dx=\left[x^{3}-\dfrac{x^{2}}{2}+x\right]_{-2}^{2}
Substitute x=2 and x=-2 and subtract. Put brackets round the negative value.
=\left(2^{3}-\dfrac{2^{2}}{2}+2\right)-\left((-2)^{3}-\dfrac{(-2)^{2}}{2}+(-2)\right)=(8-2+2)-(-8-2-2)=8+12=20

Example 25 (substitution in a definite integral)

Evaluate \int_{0}^{2}\sqrt{2x^{2}+1}\;x\,dx.

Method 1: substitute, then return to x before using the limits

Let u=2x^{2}+1, so du=4x\,dx. Multiply and divide by 4.
\int_{0}^{2}\sqrt{2x^{2}+1}\;x\,dx=\dfrac14\int_{0}^{2}\sqrt{2x^{2}+1}\;(4x\,dx)
In terms of u the integrand is \tfrac14\sqrt{u}\,du, and \int u^{1/2}du=\tfrac23u^{3/2}. Return to x before using the limits, because 0 and 2 are x-values.
\dfrac14\int\sqrt{u}\,du=\dfrac14\cdot\dfrac23u^{3/2}=\dfrac16(2x^{2}+1)^{3/2}\;\Longrightarrow\;\int_{0}^{2}\sqrt{2x^{2}+1}\;x\,dx=\left[\dfrac16(2x^{2}+1)^{3/2}\right]_{0}^{2}
Apply the limits.
=\dfrac16\left[9^{3/2}-1^{3/2}\right]=\dfrac16(27-1)=\dfrac{26}{6}=\dfrac{13}{3}

Method 2: change the limits

With u=2x^{2}+1: when x=0, u=1; when x=2, u=9. The new limits are u-values.
x=0\Rightarrow u=1,\qquad x=2\Rightarrow u=9
Write the whole integral in u with the new limits and evaluate. There is no need to return to x.
\dfrac14\int_{1}^{9}\sqrt{u}\,du=\dfrac14\cdot\dfrac23\Big[u^{3/2}\Big]_{1}^{9}=\dfrac16(27-1)=\dfrac{13}{3}

\dfrac{13}{3}

Example 26

Evaluate \displaystyle\int_{a}^{b}\frac{1}{1-\cos x}\,dx when a=\dfrac{\pi}{4} and b=\dfrac{\pi}{3}.

Multiply the numerator and the denominator by 1+\cos x, and use 1-\cos^{2}x=\sin^{2}x.
\dfrac{1}{1-\cos x}=\dfrac{1+\cos x}{1-\cos^{2}x}=\dfrac{1+\cos x}{\sin^{2}x}=\operatorname{cosec}^{2}x+\cot x\operatorname{cosec} x
Integrate: \int\operatorname{cosec}^{2}x\,dx=-\cot x and \int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x.
\int_{a}^{b}\dfrac{dx}{1-\cos x}=\Big[-\cot x-\operatorname{cosec} x\Big]_{a}^{b}
Apply the limits, using \cot\tfrac{\pi}{3}=\tfrac{1}{\sqrt3}, \operatorname{cosec}\tfrac{\pi}{3}=\tfrac{2}{\sqrt3}, \cot\tfrac{\pi}{4}=1, \operatorname{cosec}\tfrac{\pi}{4}=\sqrt2.
=-\left(\dfrac{1}{\sqrt3}-1\right)-\left(\dfrac{2}{\sqrt3}-\sqrt2\right)=1+\sqrt2-\dfrac{3}{\sqrt3}=1+\sqrt2-\sqrt3

1+\sqrt{2}-\sqrt{3}

Example 27

Evaluate \int_{1}^{e}x\ln x\,dx.

Integrate by parts with \ln x as the first function and x as the second, keeping the limits.
\int_{1}^{e}x\ln x\,dx=\left[\ln x\cdot\dfrac{x^{2}}{2}\right]_{1}^{e}-\int_{1}^{e}\dfrac1x\cdot\dfrac{x^{2}}{2}\,dx=\left[\dfrac{x^{2}\ln x}{2}\right]_{1}^{e}-\dfrac12\left[\dfrac{x^{2}}{2}\right]_{1}^{e}
Use \ln e=1 and \ln1=0.
=\left(\dfrac{e^{2}}{2}-0\right)-\dfrac12\left(\dfrac{e^{2}}{2}-\dfrac12\right)=\dfrac{e^{2}}{2}-\dfrac{e^{2}}{4}+\dfrac14=\dfrac{e^{2}}{4}+\dfrac14

\dfrac{e^{2}+1}{4}

Check points

Evaluate (i) \int_{\pi/6}^{\pi}\cos x\,dx (ii) \int_{0}^{1}\sin^{-1}x\,dx.

(i) An antiderivative of \cos x is \sin x.
\int_{\pi/6}^{\pi}\cos x\,dx=\Big[\sin x\Big]_{\pi/6}^{\pi}=\sin\pi-\sin\dfrac{\pi}{6}=0-\dfrac12=-\dfrac12
(ii) Integrate by parts with \sin^{-1}x first and 1 second; the remaining integral \int\dfrac{x}{\sqrt{1-x^{2}}}\,dx=-\sqrt{1-x^{2}}.
\int\sin^{-1}x\,dx=x\sin^{-1}x-\int\dfrac{x}{\sqrt{1-x^{2}}}\,dx=x\sin^{-1}x+\sqrt{1-x^{2}}
Apply the limits.
\int_{0}^{1}\sin^{-1}x\,dx=\Big[x\sin^{-1}x+\sqrt{1-x^{2}}\Big]_{0}^{1}=\left(\dfrac{\pi}{2}+0\right)-(0+1)=\dfrac{\pi}{2}-1

3.9, 3.10.1Area between a curve and the x-axisOften examined

The area of a rectangle is length times width, and the area of a triangle is half the base times the height. A polygon can be cut into triangles and its area found by adding them; the answer does not depend on how it is cut. A region with a curved boundary cannot be cut into triangles, so its area is defined with a definite integral. Volumes of solids are found the same way (Section 3.11).

Area of a bounded region (3.10.1). If f is continuous and f(x)\ge0 on [a,b], the area of the region under the graph of f, above the x-axis, and between the lines x=a and x=b is

A=\int_{a}^{b}f(x)\,dx

Example 28

Find the area of the region bounded by the line 2y+x=8, the x-axis, and the lines x=2 and x=4.

Make y the subject.
2y+x=8\;\Longrightarrow\;y=\dfrac{8-x}{2}=4-\dfrac{x}{2}
The line is above the x-axis for 2\le x\le4 (at x=4, y=2>0). The region is the trapezium ABCD in the graph.
A=\int_{2}^{4}\left(4-\dfrac{x}{2}\right)dx=\left[4x-\dfrac{x^{2}}{4}\right]_{2}^{4}
Apply the limits.
=\left[4(4)-\dfrac{4^{2}}{4}\right]-\left[4(2)-\dfrac{2^{2}}{4}\right]=(16-4)-(8-1)=5

Check with geometry: the trapezium has parallel sides AD=3 and BC=2 and width 2, so its area is \tfrac12(3+2)(2)=5.

A=5\ \text{sq. units}

When the curve goes below the x-axis

If f(x)<0 on part of the interval, the integral over that part is negative (Theorem 3.2: net signed area). Area is never negative, so:

find where the curve cuts the x-axis inside [a,b] by solving f(x)=0;
integrate separately over each piece;
add the absolute values: A=\left|\int_{a}^{c}f(x)\,dx\right|+\left|\int_{c}^{b}f(x)\,dx\right|.

Area across the axis

Find the area bounded by y=x^{2}-4, the x-axis, and the lines x=0 and x=3.

The curve cuts the x-axis where x^{2}-4=0, that is at x=2 inside [0,3]. It is below the axis on [0,2] and above it on [2,3].
x^{2}-4=0\;\Longrightarrow\;x=2
Integrate over [0,2]. The result is negative because this part is below the axis.
\int_{0}^{2}(x^{2}-4)\,dx=\left[\dfrac{x^{3}}{3}-4x\right]_{0}^{2}=\dfrac83-8=-\dfrac{16}{3}
Integrate over [2,3].
\int_{2}^{3}(x^{2}-4)\,dx=\left[\dfrac{x^{3}}{3}-4x\right]_{2}^{3}=(9-12)-\left(\dfrac83-8\right)=-3+\dfrac{16}{3}=\dfrac73
Add the absolute values.
A=\dfrac{16}{3}+\dfrac73=\dfrac{23}{3}

A=\dfrac{23}{3}\approx7.67\ \text{sq. units}

Area with respect to y

If a curve is given as x=g(y) with g(y)\ge0, the area between the curve, the y-axis and the lines y=c and y=d is found with horizontal strips, integrating with respect to y:

A=\int_{c}^{d}g(y)\,dy

Integrating along y

Find the area between the curve x=y^{2}, the y-axis, and the lines y=0 and y=2.

Here g(y)=y^{2}\ge0 is the horizontal distance from the y-axis to the curve.
A=\int_{0}^{2}y^{2}\,dy=\left[\dfrac{y^{3}}{3}\right]_{0}^{2}=\dfrac83

A=\dfrac{8}{3}\ \text{sq. units}

3.10.2Area between two curvesOften examined

If f(x)\ge g(x) for all x between a and b, then the area under the graph of f minus the area under the graph of g is the area between the curves:

A=\int_{a}^{b}\big[f(x)-g(x)\big]\,dx,\qquad f(x)\ge g(x)\ \text{on}\ [a,b]

The formula works even when part of the region is below the x-axis, because only the vertical distance f(x)-g(x) between the curves matters.

Find the x-coordinates of the intersection points by solving f(x)=g(x). When the region is enclosed by the two curves, these are the limits.
Decide which curve is on top: test one x-value between the limits, or use the sketch.
Integrate (top minus bottom) between the limits.
If the curves cross inside the interval, split there and add the pieces.

Example 29

Find the area of the region bounded by the graphs of f(x)=(x-1)^{2} and g(x)=3-x.

Find the intersection points.
(x-1)^{2}=3-x\;\Longrightarrow\;x^{2}-2x+1=3-x\;\Longrightarrow\;x^{2}-x-2=0\;\Longrightarrow\;(x-2)(x+1)=0
So x=-1 and x=2, giving the points (-1,4) and (2,1). Test x=0: g(0)=3>f(0)=1, so the line is on top.
g(x)\ge f(x)\quad\text{for}\quad-1\le x\le2
Integrate top minus bottom.
A=\int_{-1}^{2}\big[(3-x)-(x-1)^{2}\big]dx=\int_{-1}^{2}(2+x-x^{2})\,dx=\left[2x+\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{-1}^{2}
Apply the limits.
=\left(4+2-\dfrac83\right)-\left(-2+\dfrac12+\dfrac13\right)=\dfrac{10}{3}+\dfrac76=\dfrac92

A=\dfrac{9}{2}=4.5\ \text{sq. units}

A curve and a line

Find the area enclosed by y=x^{2} and y=2x.

Intersections: x^{2}=2x gives x(x-2)=0, so x=0 and x=2 (points (0,0) and (2,4)).
x^{2}-2x=0\;\Longrightarrow\;x=0,\ x=2
At x=1: 2x=2>x^{2}=1, so the line is on top.
A=\int_{0}^{2}(2x-x^{2})\,dx=\left[x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{2}=4-\dfrac83=\dfrac43

A=\dfrac{4}{3}\ \text{sq. units}

3.11Volume of solids of revolution (disc method)Often examined

Turn the region under y=f(x) from x=a to x=b once about the x-axis. It sweeps out a solid of revolution, like a vase on a potter's wheel. Every cross-section at right angles to the axis is a circle (a disc) of radius f(x).

Derivation The disc method (3.11.1)

Aim. Show that the volume of the solid formed by rotating the region under y=f(x), a\le x\le b, about the x-axis is V=\pi\int_{a}^{b}[f(x)]^{2}\,dx.

The region under y=f(x) from a to b rotated about the x-axis. A thin slice at x is a disc of radius r=f(x) and thickness \Delta x; its volume is \pi[f(x)]^{2}\Delta x.

Partition [a,b] as a=x_{0}<x_{1}<\dots<x_{n}=b. When the thin strip over the i-th subinterval is rotated, it makes an almost flat disc D_{i} of radius f(x_{i}) and thickness (height) \Delta x_{i}.
\text{radius}=f(x_{i}),\qquad\text{thickness}=\Delta x_{i}
A disc is a short cylinder, so its volume is area of base times height, \pi r^{2}h.
V_{i}=\pi\,[f(x_{i})]^{2}\,\Delta x_{i}
Add the volumes of all n discs.
V\approx\sum_{i=1}^{n}\pi\,[f(x_{i})]^{2}\,\Delta x_{i}
Let \Delta x_{i}\to0. The sum becomes a definite integral (Definition 3.2).
V=\lim_{\Delta x_{i}\to0}\sum_{i=1}^{n}\pi[f(x_{i})]^{2}\Delta x_{i}=\pi\int_{a}^{b}[f(x)]^{2}\,dx

ResultV=\pi\int_{a}^{b}[f(x)]^{2}\,dx=\pi\int_{a}^{b}y^{2}\,dx

Square the radius, multiply by \pi, and integrate along the axis of rotation. The answer is in cubic units.

About the y-axis. If the region between the curve x=g(y) and the y-axis, from y=c to y=d, is rotated about the y-axis, the discs are horizontal, with radius g(y) and thickness \Delta y:

V=\pi\int_{c}^{d}[g(y)]^{2}\,dy=\pi\int_{c}^{d}x^{2}\,dy

Example 30

Find the volume of the solid obtained by rotating the graph of y=x^{2} between x=1 and x=2 about the x-axis.

The radius of the disc at x is f(x)=x^{2}.
V=\pi\int_{1}^{2}[f(x)]^{2}\,dx=\pi\int_{1}^{2}(x^{2})^{2}\,dx=\pi\int_{1}^{2}x^{4}\,dx
Integrate and apply the limits.
V=\pi\left[\dfrac{x^{5}}{5}\right]_{1}^{2}=\dfrac{\pi}{5}\left(2^{5}-1^{5}\right)=\dfrac{31\pi}{5}

The shaded region is rotated about the x-axis. Its mirror image (dashed) shows the outline of the solid, which is widest at x=2.

V=\dfrac{31\pi}{5}\approx19.5\ \text{cubic units}

Example 31

Find the volume of the solid generated when the region enclosed by y=\sqrt{x}, y=0 and y=2 is revolved about the y-axis.

The discs are perpendicular to the y-axis, so write x in terms of y: y=\sqrt{x} gives x=y^{2}. The radius of the disc at height y is g(y)=y^{2}.
g(y)=y^{2},\qquad0\le y\le2
Use the y-axis formula.
V=\pi\int_{0}^{2}[g(y)]^{2}\,dy=\pi\int_{0}^{2}(y^{2})^{2}\,dy=\pi\int_{0}^{2}y^{4}\,dy
Integrate and apply the limits.
V=\pi\left[\dfrac{y^{5}}{5}\right]_{0}^{2}=\dfrac{\pi}{5}\left(2^{5}-0^{5}\right)=\dfrac{32\pi}{5}

The region between the y-axis and x=y^{2}, for 0\le y\le2, is rotated about the y-axis. Each horizontal slice is a disc of radius y^{2}.

V=\dfrac{32\pi}{5}\approx20.1\ \text{cubic units}

Volume of a cone

Use the disc method to show that a right circular cone of base radius r and height h has volume \tfrac13\pi r^{2}h.

Put the vertex at the origin and the axis along the x-axis. The slant edge is the line through (0,0) and (h,r), so y=\dfrac{r}{h}x for 0\le x\le h. Rotating the triangle under this line about the x-axis gives the cone.
f(x)=\dfrac{r}{h}x
Apply the disc formula; r and h are constants.
V=\pi\int_{0}^{h}\left(\dfrac{r}{h}x\right)^{2}dx=\dfrac{\pi r^{2}}{h^{2}}\left[\dfrac{x^{3}}{3}\right]_{0}^{h}=\dfrac{\pi r^{2}}{h^{2}}\cdot\dfrac{h^{3}}{3}=\dfrac13\pi r^{2}h

V=\dfrac{1}{3}\pi r^{2}h

Average value of a functionOften examined

The average of n numbers is their sum divided by n. A continuous function takes infinitely many values on [a,b], so we replace the sum by an integral.

Derivation Average value formula

Aim. Show that the average value of a continuous function f on [a,b] is \dfrac{1}{b-a}\displaystyle\int_{a}^{b}f(x)\,dx.

Take a regular partition of [a,b] into n pieces of width \Delta x=\dfrac{b-a}{n} and sample f once in each piece. The average of the n sample values is their sum divided by n.
\text{average}\approx\dfrac{f(x_{1})+f(x_{2})+\dots+f(x_{n})}{n}
Since \Delta x=\dfrac{b-a}{n}, we have \dfrac1n=\dfrac{\Delta x}{b-a}.
\text{average}\approx\dfrac{1}{b-a}\sum_{k=1}^{n}f(x_{k})\,\Delta x
Let n\to\infty; the sum becomes the definite integral.
f_{\text{avg}}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx

Resultf_{\text{avg}}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx

The rectangle on [a,b] with height f_{\text{avg}} has the same area as the region under the curve. A continuous f takes the value f_{\text{avg}} at least once in [a,b] (the mean value theorem for integrals).

The dashed rectangle (width 3, height 3) has area 9, the same as the shaded area under y=x^{2} on [0,3]. The curve reaches the average height 3 at x=c=\sqrt3.

Average value of x²

Find the average value of f(x)=x^{2} on (i) [0,3] (ii) [1,3].

(i) Here b-a=3.
f_{\text{avg}}=\dfrac{1}{3-0}\int_{0}^{3}x^{2}\,dx=\dfrac13\left[\dfrac{x^{3}}{3}\right]_{0}^{3}=\dfrac13(9)=3
(ii) Here b-a=2.
f_{\text{avg}}=\dfrac{1}{3-1}\int_{1}^{3}x^{2}\,dx=\dfrac12\left[\dfrac{x^{3}}{3}\right]_{1}^{3}=\dfrac12\left(9-\dfrac13\right)=\dfrac12\cdot\dfrac{26}{3}=\dfrac{13}{3}

Average value of sin x

Find the average value of \sin x on [0,\pi].

Use the formula with b-a=\pi.
f_{\text{avg}}=\dfrac1\pi\int_{0}^{\pi}\sin x\,dx=\dfrac1\pi\Big[-\cos x\Big]_{0}^{\pi}=\dfrac1\pi(1+1)=\dfrac{2}{\pi}\approx0.64

\dfrac{2}{\pi}

Trapezium rule

When an antiderivative is hard to find, or f is known only from a table of values, we estimate \int_{a}^{b}f(x)\,dx numerically. The trapezium rule replaces the curve over each strip by a straight line, so every strip becomes a trapezium.

Derivation The trapezium rule

Aim. Estimate \int_{a}^{b}f(x)\,dx using n strips of equal width.

Four trapezia under y=x^{2} on [0,2] with h=0.5. Each straight top lies slightly above the curve, so the rule over-estimates here.

Divide [a,b] into n equal strips of width h. Let y_{0},y_{1},\dots,y_{n} be the values of f at x_{0}=a,\ x_{1}=a+h,\ \dots,\ x_{n}=b.
h=\dfrac{b-a}{n},\qquad y_{k}=f(x_{k})
The k-th strip is a trapezium with parallel sides y_{k-1} and y_{k} and width h.
\text{area of strip }k=\dfrac{h}{2}\big(y_{k-1}+y_{k}\big)
Add all the strips. Every inner ordinate y_{1},\dots,y_{n-1} belongs to two neighbouring strips, so it is counted twice.
\int_{a}^{b}f(x)\,dx\approx\dfrac{h}{2}\Big[y_{0}+2\big(y_{1}+y_{2}+\dots+y_{n-1}\big)+y_{n}\Big]

Result\int_{a}^{b}f(x)\,dx\approx\dfrac{h}{2}\Big[y_{0}+2\big(y_{1}+\dots+y_{n-1}\big)+y_{n}\Big],\qquad h=\dfrac{b-a}{n}

First and last ordinates once, all the others twice, then multiply by half the strip width. More strips give a better estimate.

Trapezium rule with 4 strips

Use the trapezium rule with n=4 to estimate \int_{0}^{2}x^{2}\,dx, and compare with the exact value.

Find the strip width and the ordinates.
h=\dfrac{2-0}{4}=0.5;\qquad y_{0},y_{1},y_{2},y_{3},y_{4}=0,\ 0.25,\ 1,\ 2.25,\ 4
Apply the rule.
T=\dfrac{0.5}{2}\Big[0+2(0.25+1+2.25)+4\Big]=0.25\,[7+4]=2.75
Exact value by the Fundamental Theorem.
\int_{0}^{2}x^{2}\,dx=\left[\dfrac{x^{3}}{3}\right]_{0}^{2}=\dfrac83\approx2.667
The estimate is slightly too large because the curve bends upwards. With a single strip (n=1) the rule gives \tfrac{2}{2}(0+4)=4, a much worse estimate.

3.12, 3.12.1Consumer and producer surplus

Economists use definite integrals to define consumer and producer surplus. Let a commodity sell at a price x per unit. Let D(x) be the number of units demanded by consumers and S(x) the number of units supplied by manufacturers at that price. Demand falls as the price rises; supply rises as the price rises.

Equilibrium price: The price p at which demand equals supply, D(p)=S(p). The market is then said to be in equilibrium.
Consumer surplus: C_{s}=\displaystyle\int_{p}^{b}D(x)\,dx, where b is the price at which the demand becomes zero, D(b)=0.
Producer surplus: P_{s}=\displaystyle\int_{c}^{p}S(x)\,dx, where c is the price at which the supply is zero, S(c)=0.

Example 32

The demand and supply of a commodity selling for x dollars a unit are D(x)=1000-20x and S(x)=x^{2}+10x. Find the consumer and producer surplus.

Equilibrium price: solve D(x)=S(x).
1000-20x=x^{2}+10x\;\Longrightarrow\;x^{2}+30x-1000=0\;\Longrightarrow\;(x+50)(x-20)=0\;\Longrightarrow\;p=20
Demand is zero at b: 1000-20b=0 gives b=50. Supply is zero at c: c^{2}+10c=0 gives c=0 (a price cannot be negative).
b=50,\qquad c=0
Consumer surplus: the area under D(x) on [20,50].
C_{s}=\int_{20}^{50}(1000-20x)\,dx=\Big[1000x-10x^{2}\Big]_{20}^{50}=(50000-25000)-(20000-4000)=9000
Producer surplus: the area under S(x) on [0,20].
P_{s}=\int_{0}^{20}(x^{2}+10x)\,dx=\left[\dfrac{x^{3}}{3}+5x^{2}\right]_{0}^{20}=\dfrac{8000}{3}+2000\approx4666.67

C_{s} is the area under the demand line from the equilibrium price p=20 to b=50; P_{s} is the area under the supply curve from c=0 to p=20.

C_{s}=\$\,9000,\qquad P_{s}\approx\$\,4666.67

3.12.2Rectilinear motion

If S(t) is the position at time t of an object moving along a straight line, its velocity and acceleration are derivatives:

v(t)=\dfrac{dS}{dt},\qquad a(t)=\dfrac{dv}{dt}

Integration reverses this, so position and velocity are indefinite integrals:

S(t)=\int v(t)\,dt,\qquad v(t)=\int a(t)\,dt

Each integration brings in a constant. The constants are found from the initial position S(0) and the initial velocity v(0).

Key facts from the book. The sign of g depends on which direction is taken as positive.
Motion	Initial position	Initial velocity	Acceleration
upward motion (from the ground)	S(0)=0	v(0)>0	a=g=-9.8\,\mathrm{m\,s^{-2}}=-32\,\mathrm{ft\,s^{-2}}
downward motion (dropped from height h)	S(0)=h	v(0)=0	a=g=9.8\,\mathrm{m\,s^{-2}}=32\,\mathrm{ft\,s^{-2}}

Displacement and distance

Over a time interval [t_{1},t_{2}] the displacement (change in position) is \int_{t_{1}}^{t_{2}}v(t)\,dt=S(t_{2})-S(t_{1}). The distance travelled is \int_{t_{1}}^{t_{2}}|v(t)|\,dt: if the velocity changes sign, split the interval where v=0 and add the absolute values.

Example 33

The position function of an object that moves on a coordinate line is S(t)=t^{2}-6t, where S is measured in centimetres and t in seconds. Find the distance travelled in the time interval [3,9].

Velocity is the derivative of position.
v(t)=\dfrac{dS}{dt}=2t-6
For 3\le t\le9, v(t)=2t-6\ge0, so the object never turns back and the distance equals the displacement.
\text{distance}=\int_{3}^{9}(2t-6)\,dt
Evaluate.
=\Big[t^{2}-6t\Big]_{3}^{9}=(81-54)-(9-18)=27+9=36\ \mathrm{cm}

\text{distance}=36\ \mathrm{cm}

Velocity against time for S(t)=t^{2}-6t

Displacement is the net area under the velocity graph; distance is the total area. On [3,9] both equal 36. On [0,9] the displacement is 36-9=27 but the distance is 36+9=45.

Distance when the object turns back

For the same object, S(t)=t^{2}-6t, find the displacement and the distance travelled in [0,9].

v(t)=2t-6 is negative for t<3 and positive for t>3: the object moves backwards, stops at t=3, then moves forwards.
v(t)=0\;\Longrightarrow\;t=3
Displacement: integrate v over the whole interval.
\int_{0}^{9}(2t-6)\,dt=\Big[t^{2}-6t\Big]_{0}^{9}=27-0=27\ \mathrm{cm}
Distance: split at t=3 and add the absolute values.
\left|\int_{0}^{3}(2t-6)\,dt\right|+\int_{3}^{9}(2t-6)\,dt=|-9|+36=45\ \mathrm{cm}

Using initial conditions

A ball is thrown vertically upward from the ground with speed 49\,\mathrm{m\,s^{-1}}. Taking a=-9.8\,\mathrm{m\,s^{-2}}, find v(t) and S(t), the time to reach the highest point, and the maximum height.

Integrate the acceleration, and use v(0)=49 to find the constant.
\begin{aligned}v(t)&=\int(-9.8)\,dt=-9.8t+c_{1}\\ v(0)&=49\Rightarrow c_{1}=49\\ v(t)&=49-9.8t\end{aligned}
Integrate the velocity, and use S(0)=0 (it starts on the ground).
S(t)=\int(49-9.8t)\,dt=49t-4.9t^{2}+c_{2},\qquad S(0)=0\Rightarrow c_{2}=0
At the highest point the velocity is zero.
49-9.8t=0\;\Longrightarrow\;t=5\ \mathrm{s}
Maximum height.
S(5)=49(5)-4.9(25)=245-122.5=122.5\ \mathrm{m}

3.12.3, 3.12.4Work done by a variable force and springs

When a constant force F moves an object a distance d in its own direction, the work done is W=Fd. If the force changes with position, cut the path into small pieces. On a piece of length \Delta x_{k} the force is nearly constant, so the work is about F(x_{k})\Delta x_{k}. Adding the pieces and taking the limit gives an integral.

Work done by a variable force: If a continuous force F(x) acts along the x-axis, the work done in moving an object from x=a to x=b is W=\displaystyle\int_{a}^{b}F(x)\,dx. With F in newtons and x in metres, W is in joules.

Motion of a spring (3.12.4)

Hooke's law: when a spring is stretched (or compressed) beyond its natural length, the restoring force exerted by the spring is directly proportional to the extension (or compression). So, to stretch a spring x units beyond its natural length we must apply the force

F(x)=kx

k is the spring constant, in \mathrm{N\,m^{-1}}. A stiffer spring has a larger k.

A spring at its natural length (top) and stretched by an extension x (bottom). The force needed to hold it is F=kx, where x is measured from the natural length.

W=\int_{0}^{d}kx\,dx=\left[\dfrac{kx^{2}}{2}\right]_{0}^{d}=\dfrac12kd^{2}

Work done in stretching a spring by d from its natural length.

Example 34

A force of 130\,\mathrm{N} is required to stretch a spring 50\,\mathrm{cm}. Find the work done in stretching the spring 20\,\mathrm{cm} beyond its natural (unstretched) length.

Convert to metres and find k from F=kx.
x=50\,\mathrm{cm}=0.5\,\mathrm{m},\quad F=130\,\mathrm{N}:\qquad130=k(0.5)\;\Longrightarrow\;k=260\,\mathrm{N\,m^{-1}}
So the force law is F=260x. Stretching by 20\,\mathrm{cm}=0.2\,\mathrm{m}=\tfrac15\,\mathrm{m} from the natural length:
W=\int_{0}^{0.2}260x\,dx=\Big[130x^{2}\Big]_{0}^{0.2}=130(0.04)=5.2\,\mathrm{J}

W=5.2\,\mathrm{J}

3.12Other applications: total change from a rate

The unit's learning outcomes also mention the volume of a container, the growth rate of a population, the time for an investment to give a return, and drug dosage. All of them use one idea from the Fundamental Theorem: integrating a rate of change gives the total change.

Q(b)-Q(a)=\int_{a}^{b}Q'(t)\,dt

Q'(t) is the rate at which the quantity Q changes; the integral adds up all the small changes Q'(t)\,\Delta t.

Situation	Rate given	The integral gives
population	growth rate P'(t) (people per year)	increase in population
sales or investment	rate of revenue R'(t) (dollars per year)	total revenue or return
medicine	rate at which a drug enters the blood (mg per hour)	total dose absorbed
container	rate of flow r(t) (litres per minute)	volume of liquid added

Population growth

The population of a town grows at the rate P'(t)=400+60t people per year, where t is in years. By how much does the population increase in the first 5 years?

Integrate the growth rate from t=0 to t=5.
\int_{0}^{5}(400+60t)\,dt=\Big[400t+30t^{2}\Big]_{0}^{5}=2000+750=2750

2750\ \text{people}

Time for an investment

An investment earns money at the rate R'(t)=2000\sqrt{t} dollars per year. How long does it take to earn \$\,18000 in total?

Total earned in the first T years.
\int_{0}^{T}2000t^{1/2}\,dt=2000\cdot\dfrac23\Big[t^{3/2}\Big]_{0}^{T}=\dfrac{4000}{3}T^{3/2}
Set this equal to 18000 and solve for T.
\dfrac{4000}{3}T^{3/2}=18000\;\Longrightarrow\;T^{3/2}=13.5\;\Longrightarrow\;T=13.5^{2/3}\approx5.67\ \text{years}

Drug dosage

A drug enters a patient's bloodstream at the rate r(t)=20e^{-0.5t} mg per hour. How much of the drug enters in the first 4 hours?

Integrate the rate, using \int e^{kt}dt=\dfrac{e^{kt}}{k} with k=-0.5.
\int_{0}^{4}20e^{-0.5t}\,dt=20\left[\dfrac{e^{-0.5t}}{-0.5}\right]_{0}^{4}=-40\left(e^{-2}-1\right)=40\left(1-e^{-2}\right)\approx34.6\ \mathrm{mg}

Volume of a container

A bowl is made by rotating the curve y=x^{2}, 0\le y\le4, about the y-axis (lengths in cm). Find the volume of liquid it holds when full.

About the y-axis the radius at height y is x=\sqrt{y}, so x^{2}=y.
V=\pi\int_{0}^{4}x^{2}\,dy=\pi\int_{0}^{4}y\,dy=\pi\left[\dfrac{y^{2}}{2}\right]_{0}^{4}=8\pi\approx25.1\ \mathrm{cm^{3}}

Formula sheet

Power rule (Formula 3.1)

\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+c,\qquad n\neq-1

n: any real number except -1; c: constant of integration

Reciprocal (Formula 3.2)

\int\dfrac{1}{x}\,dx=\ln|x|+c

x\neq0

Exponential (Formula 3.3)

\int e^{x}\,dx=e^{x}+c

Exponential with base a (Formula 3.4)

\int a^{x}\,dx=\dfrac{a^{x}}{\ln a}+c

a>0, a\neq1

General power rule

\int\big[f(x)\big]^{n}f'(x)\,dx=\dfrac{\big[f(x)\big]^{n+1}}{n+1}+c,\qquad n\neq-1

the derivative of the bracket must be a factor

Derivative over function

\int\dfrac{f'(x)}{f(x)}\,dx=\ln\big|f(x)\big|+c

General exponential forms

\int e^{f(x)}f'(x)\,dx=e^{f(x)}+c,\qquad\int a^{f(x)}f'(x)\,dx=\dfrac{a^{f(x)}}{\ln a}+c

Linear inner function

\begin{aligned}\int(ax+b)^{n}\,dx&=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c\\ \int\dfrac{dx}{ax+b}&=\dfrac1a\ln|ax+b|+c\\ \int e^{ax+b}\,dx&=\dfrac1ae^{ax+b}+c\end{aligned}

a\neq0; integrate as usual and divide by a

Linearity (Theorem 3.1)

\int\big[af(x)\pm bg(x)\big]\,dx=a\int f(x)\,dx\pm b\int g(x)\,dx

a, b: constants

Sine and cosine

\int\sin kx\,dx=-\dfrac{\cos kx}{k}+c,\qquad\int\cos kx\,dx=\dfrac{\sin kx}{k}+c

k\neq0; angles in radians

Squares of secant and cosecant

\int\sec^{2}x\,dx=\tan x+c,\qquad\int\operatorname{cosec}^{2}x\,dx=-\cot x+c

Secant times tangent, cosecant times cotangent

\int\sec x\tan x\,dx=\sec x+c,\qquad\int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x+c

Tangent and cotangent

Secant and cosecant

\begin{aligned}\int\sec x\,dx&=\ln|\sec x+\tan x|+c\\ \int\operatorname{cosec} x\,dx&=\ln|\operatorname{cosec} x-\cot x|+c\end{aligned}

Squares of sine and cosine

\int\sin^{2}x\,dx=\dfrac{x}{2}-\dfrac{\sin2x}{4}+c,\qquad\int\cos^{2}x\,dx=\dfrac{x}{2}+\dfrac{\sin2x}{4}+c

from \sin^{2}x=\tfrac12(1-\cos2x) and \cos^{2}x=\tfrac12(1+\cos2x)

Squares of tangent and cotangent

\int\tan^{2}x\,dx=\tan x-x+c,\qquad\int\cot^{2}x\,dx=-\cot x-x+c

from \tan^{2}x=\sec^{2}x-1 and \cot^{2}x=\operatorname{cosec}^{2}x-1

Substitution

\int f\big(g(x)\big)\,g'(x)\,dx=\int f(t)\,dt,\qquad t=g(x)

dt=g'(x)\,dx; substitute back at the end

Formula 3.5

\int\dfrac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\dfrac{x}{a}\right)+c

a>0; substitution x=a\sin\theta

Formula 3.6

\int\sqrt{a^{2}-x^{2}}\,dx=\dfrac{a^{2}}{2}\sin^{-1}\left(\dfrac{x}{a}\right)+\dfrac{x\sqrt{a^{2}-x^{2}}}{2}+c

substitution x=a\sin\theta, or by parts

Formula 3.7

\int\dfrac{dx}{\sqrt{x^{2}-a^{2}}}=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+c

substitution x=a\sec\theta

Formula 3.8

\int\dfrac{dx}{\sqrt{x^{2}+a^{2}}}=\ln\left(x+\sqrt{x^{2}+a^{2}}\right)+c

substitution x=a\tan\theta

Formula 3.9

\int\dfrac{dx}{a^{2}+x^{2}}=\dfrac1a\tan^{-1}\left(\dfrac{x}{a}\right)+c

substitution x=a\tan\theta

Roots by parts

\int\sqrt{a^{2}+x^{2}}\,dx=\dfrac{x\sqrt{a^{2}+x^{2}}}{2}+\dfrac{a^{2}}{2}\ln\left(x+\sqrt{a^{2}+x^{2}}\right)+c

Example 15 (ii), corrected

Roots by parts (check point)

\int\sqrt{x^{2}-a^{2}}\,dx=\dfrac{x\sqrt{x^{2}-a^{2}}}{2}-\dfrac{a^{2}}{2}\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+c

equivalently -\dfrac{a^{2}}{2}\cosh^{-1}\dfrac{x}{a}

Difference of squares

\int\dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right|+c

from partial fractions

Integration by parts

\int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int\left[f'(x)\int g(x)\,dx\right]dx

f: first function (differentiated); g: second function (integrated)

Exponential times sine or cosine

\begin{aligned}\int e^{ax}\sin bx\,dx&=\dfrac{e^{ax}(a\sin bx-b\cos bx)}{a^{2}+b^{2}}+c\\ \int e^{ax}\cos bx\,dx&=\dfrac{e^{ax}(a\cos bx+b\sin bx)}{a^{2}+b^{2}}+c\end{aligned}

Definite integral (Definition 3.2)

\int_{a}^{b}f(x)\,dx=\lim_{\Delta x\to0}\sum_{k=1}^{n}f(x_{k})\,\Delta x_{k}

a: lower limit; b: upper limit; regular partition \Delta x=\dfrac{b-a}{n}

Fundamental theorem of calculus

\int_{a}^{b}f(x)\,dx=\Big[F(x)\Big]_{a}^{b}=F(b)-F(a)

F: any antiderivative of f

Derivative of an integral

\dfrac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x)

Properties of the definite integral

\int_{a}^{a}f\,dx=0,\qquad\int_{b}^{a}f\,dx=-\int_{a}^{b}f\,dx,\qquad\int_{a}^{b}f\,dx=\int_{a}^{c}f\,dx+\int_{c}^{b}f\,dx

Even and odd functions

\int_{-a}^{a}f(x)\,dx=\begin{cases}2\displaystyle\int_{0}^{a}f(x)\,dx, & f\ \text{even}\\[4pt] 0, & f\ \text{odd}\end{cases}

Area under a curve

A=\int_{a}^{b}f(x)\,dx

f(x)\ge0 on [a,b]; if f changes sign, split at the x-intercepts and add absolute values

square units

Area with respect to y

A=\int_{c}^{d}g(y)\,dy

curve x=g(y)\ge0 between y=c and y=d

square units

Area between two curves

A=\int_{a}^{b}\big[f(x)-g(x)\big]\,dx

f(x)\ge g(x) on [a,b]; a, b from f(x)=g(x)

square units

Volume about the x-axis (disc method)

V=\pi\int_{a}^{b}\big[f(x)\big]^{2}\,dx

disc radius f(x), thickness dx

cubic units

Volume about the y-axis

V=\pi\int_{c}^{d}\big[g(y)\big]^{2}\,dy

disc radius x=g(y), thickness dy

cubic units

Average value

f_{\text{avg}}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx

Trapezium rule

\int_{a}^{b}f(x)\,dx\approx\dfrac{h}{2}\Big[y_{0}+2\big(y_{1}+\dots+y_{n-1}\big)+y_{n}\Big]

h=\dfrac{b-a}{n}, y_{k}=f(a+kh)

Consumer surplus

C_{s}=\int_{p}^{b}D(x)\,dx

x: price per unit; D: demand; p: equilibrium price (D(p)=S(p)); D(b)=0

Producer surplus

P_{s}=\int_{c}^{p}S(x)\,dx

S: supply; S(c)=0

Rectilinear motion

v=\dfrac{dS}{dt},\quad a=\dfrac{dv}{dt},\quad S(t)=\int v(t)\,dt,\quad v(t)=\int a(t)\,dt

S: position; v: velocity; a: acceleration; distance =\int|v|\,dt

Units: \mathrm{m}, \mathrm{m\,s^{-1}}, \mathrm{m\,s^{-2}}

Work done by a variable force

W=\int_{a}^{b}F(x)\,dx

F(x): force at position x

Units: joule (\mathrm{J}) with F in \mathrm{N} and x in \mathrm{m}

Hooke's law and spring work

F=kx,\qquad W=\int_{0}^{d}kx\,dx=\dfrac12kd^{2}

k: spring constant; x: extension from the natural length

Units: k in \mathrm{N\,m^{-1}}, W in \mathrm{J}

Total change from a rate

Q(b)-Q(a)=\int_{a}^{b}Q'(t)\,dt

Q'(t): rate of change of Q

Key terms

Antiderivative: A function F with F'(x)=f(x) on an interval; F is an antiderivative of f.
Indefinite integral: The family of all antiderivatives, \int f(x)\,dx=F(x)+c.
Integrand: The function f(x) being integrated.
Constant of integration: The arbitrary constant c added to every indefinite integral.
Variable of integration: The variable after d, as in dx; it shows which variable we integrate with respect to.
Integration by substitution: Changing the variable with t=g(x), dt=g'(x)\,dx, so that \int f(g(x))g'(x)\,dx=\int f(t)\,dt.
Trigonometric substitution: Putting x=a\sin\theta, a\sec\theta or a\tan\theta to remove \sqrt{a^{2}-x^{2}}, \sqrt{x^{2}-a^{2}} or \sqrt{x^{2}+a^{2}}.
Integration by parts: The rule \int fg\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx for a product of two functions.
First and second function: In integration by parts, the first function is differentiated and the second function is integrated.
Partial fractions: The simpler fractions into which a proper rational function is split before integrating.
Proper rational fraction: \dfrac{P(x)}{Q(x)} with the degree of P less than the degree of Q; otherwise the fraction is improper and must be divided first.
Partition: Points a=x_{0}<x_{1}<\dots<x_{n}=b dividing [a,b] into subintervals; regular if all have length \dfrac{b-a}{n}.
Riemann sum: \sum f(x_{k})\Delta x_{k}, the total area of rectangles that approximates the area under a curve.
Definite integral: \int_{a}^{b}f(x)\,dx, the limit of Riemann sums as \Delta x\to0; a number.
Limits of integration: The numbers a (lower limit) and b (upper limit) in \int_{a}^{b}f(x)\,dx.
Value of the integral: The number a definite integral equals.
Integrable function: A function whose Riemann sums have a limit that does not depend on the partition or the chosen points; every continuous function on [a,b] is integrable.
Net signed area: Area above the x-axis minus area below it; this is what \int_{a}^{b}f(x)\,dx measures.
Fundamental theorem of calculus: If F'=f and f is continuous on [a,b], then \int_{a}^{b}f(x)\,dx=F(b)-F(a).
Even and odd function: f is even if f(-x)=f(x) and odd if f(-x)=-f(x); then \int_{-a}^{a}f=2\int_{0}^{a}f or 0 respectively.
Solid of revolution: The solid swept out when a plane region is rotated once about an axis.
Disc method: Finding a volume of revolution by adding thin discs: V=\pi\int[f(x)]^{2}dx.
Average value: f_{\text{avg}}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx, the height of the rectangle on [a,b] with the same area as the region under f.
Trapezium rule: A numerical estimate of \int_{a}^{b}f(x)\,dx that replaces the curve on each strip by a straight line.
Equilibrium price: The price p at which demand equals supply, D(p)=S(p).
Consumer surplus: C_{s}=\int_{p}^{b}D(x)\,dx, where D(b)=0.
Producer surplus: P_{s}=\int_{c}^{p}S(x)\,dx, where S(c)=0.
Displacement and distance: Displacement is \int v\,dt (net change in position); distance travelled is \int|v|\,dt.
Hooke's law: The force needed to stretch or compress a spring is proportional to the extension: F=kx, where k is the spring constant.
Work done by a variable force: W=\int_{a}^{b}F(x)\,dx.

Exercise 3.1p. 93–94

Evaluate the following integrals. Rewrite each integrand as a sum of standard forms (powers, \dfrac1x, exponentials), or spot a function together with its derivative, then integrate term by term. Every answer has been checked by differentiating it back.

Evaluate \int(x^{2}-3x+9)\,dx.

Solution

Integrate term by term (Theorem 3.1) and take the constants 3 and 9 outside.
\int(x^{2}-3x+9)\,dx=\int x^{2}\,dx-3\int x\,dx+9\int dx
Use the power rule \int x^{n}dx=\dfrac{x^{n+1}}{n+1}; for the constant term, \int dx=x.
=\dfrac{x^{3}}{3}-3\cdot\dfrac{x^{2}}{2}+9x+c

\dfrac{x^{3}}{3}-\dfrac{3x^{2}}{2}+9x+c

Evaluate \int(y^{2}+8y+\sqrt2)\,dy.

Solution

Integrate term by term with respect to y. \sqrt2 is a constant, so its integral is \sqrt2\,y.
\int(y^{2}+8y+\sqrt2)\,dy=\int y^{2}\,dy+8\int y\,dy+\sqrt2\int dy
Apply the power rule.
=\dfrac{y^{3}}{3}+8\cdot\dfrac{y^{2}}{2}+\sqrt2\,y+c

\dfrac{y^{3}}{3}+4y^{2}+\sqrt{2}\,y+c

Evaluate \displaystyle\int\left(\sqrt{y}+\frac{1}{y^{2}}\right)dy.

Solution

Write the root and the fraction as powers of y.
\sqrt y+\dfrac{1}{y^{2}}=y^{1/2}+y^{-2}
Apply the power rule to each term: add 1 to the power and divide by the new power.
\int\left(y^{1/2}+y^{-2}\right)dy=\dfrac{y^{3/2}}{3/2}+\dfrac{y^{-1}}{-1}+c
Simplify.
=\dfrac23y^{3/2}-\dfrac1y+c

\dfrac{2}{3}y^{3/2}-\dfrac{1}{y}+c

Evaluate \int(4+x^{2})^{2}\,dx.

Solution

There is no product rule for integrals, so expand the square first, using (a+b)^{2}=a^{2}+2ab+b^{2}.
(4+x^{2})^{2}=16+8x^{2}+x^{4}
Integrate term by term with the power rule.
\int(16+8x^{2}+x^{4})\,dx=16x+\dfrac{8x^{3}}{3}+\dfrac{x^{5}}{5}+c

16x+\dfrac{8x^{3}}{3}+\dfrac{x^{5}}{5}+c

Evaluate \int(1+x)(1-x^{2})\,dx.

Solution

Multiply out the brackets first.
(1+x)(1-x^{2})=1-x^{2}+x-x^{3}
Integrate each term with the power rule.
\int(1+x-x^{2}-x^{3})\,dx=x+\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+c

x+\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+c

Evaluate \displaystyle\int\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)dx.

Solution

Write both terms as powers of x.
\sqrt x+\dfrac{1}{2\sqrt x}=x^{1/2}+\dfrac12x^{-1/2}
Apply the power rule to each term.
\int\left(x^{1/2}+\dfrac12x^{-1/2}\right)dx=\dfrac{x^{3/2}}{3/2}+\dfrac12\cdot\dfrac{x^{1/2}}{1/2}+c
Simplify. The second term makes sense: \dfrac{1}{2\sqrt x} is the derivative of \sqrt x.
=\dfrac23x^{3/2}+\sqrt x+c

\dfrac{2}{3}x^{3/2}+\sqrt{x}+c

Evaluate \int(e^{4x}-e^{-1}+1)\,dx.

Solution

e^{-1}=\dfrac1e contains no x, so it is a constant. The last two terms together are the constant 1-e^{-1}.
\int(e^{4x}-e^{-1}+1)\,dx=\int e^{4x}\,dx+\int(1-e^{-1})\,dx
Use \int e^{ax}dx=\dfrac{e^{ax}}{a} with a=4, and \int k\,dx=kx for the constant.
=\dfrac{e^{4x}}{4}+(1-e^{-1})\,x+c

\dfrac{e^{4x}}{4}+\left(1-\dfrac{1}{e}\right)x+c

Evaluate \displaystyle\int\left(e^{\frac{9}{2}x}+\frac{1}{x}\right)dx.

Solution

Split into two standard integrals.
\int\left(e^{\frac92x}+\dfrac1x\right)dx=\int e^{\frac92x}\,dx+\int\dfrac{1}{x}\,dx
Use \int e^{ax}dx=\dfrac{e^{ax}}{a} with a=\tfrac92 (dividing by \tfrac92 is the same as multiplying by \tfrac29), and \int\dfrac{dx}{x}=\ln|x|.
=\dfrac29e^{\frac92x}+\ln|x|+c

\dfrac{2}{9}e^{\frac{9}{2}x}+\ln|x|+c

92026 Annual I, Old syllabus (Curriculum 2006), MCQ 8, old syllabus

Evaluate \int xe^{x^{2}}\,dx.

Solution

The power of e is f(x)=x^{2}, with f'(x)=2x. The integrand has only x, so multiply and divide by 2.
\int xe^{x^{2}}\,dx=\dfrac12\int e^{x^{2}}(2x)\,dx
Now the integrand is e^{f(x)}f'(x), so use \int e^{f(x)}f'(x)\,dx=e^{f(x)}+c (or substitute t=x^{2}).
=\dfrac12e^{x^{2}}+c

\dfrac{1}{2}e^{x^{2}}+c

Evaluate \int5^{x}\,dx.

Solution

Use Formula 3.4, \int a^{x}dx=\dfrac{a^{x}}{\ln a}+c, with a=5.
\int5^{x}\,dx=\dfrac{5^{x}}{\ln5}+c
Check by differentiating: \dfrac{d}{dx}5^{x}=5^{x}\ln5.
\dfrac{d}{dx}\left(\dfrac{5^{x}}{\ln5}\right)=\dfrac{5^{x}\ln5}{\ln5}=5^{x}\ \checkmark

\dfrac{5^{x}}{\ln 5}+c

Evaluate \int7^{7y}\,dy.

Solution

The power is f(y)=7y, with f'(y)=7. Multiply and divide by 7.
\int7^{7y}\,dy=\dfrac17\int7^{7y}(7)\,dy
Use \int a^{f(y)}f'(y)\,dy=\dfrac{a^{f(y)}}{\ln a}+c with a=7.
=\dfrac17\cdot\dfrac{7^{7y}}{\ln7}+c=\dfrac{7^{7y}}{7\ln7}+c

\dfrac{7^{7y}}{7\ln 7}+c

Evaluate \displaystyle\int\left(x^{3}+\frac{1}{2x}-\frac{1}{x^{3}}\right)dx.

Solution

Write each term in a form that has a standard integral.
x^{3}+\dfrac{1}{2x}-\dfrac{1}{x^{3}}=x^{3}+\dfrac12\cdot\dfrac1x-x^{-3}
Use the power rule for x^{3} and x^{-3}, and \int\dfrac{dx}{x}=\ln|x| for the middle term.
\int\left(x^{3}+\dfrac12\cdot\dfrac1x-x^{-3}\right)dx=\dfrac{x^{4}}{4}+\dfrac12\ln|x|-\dfrac{x^{-2}}{-2}+c
Simplify: -\dfrac{x^{-2}}{-2}=\dfrac{1}{2x^{2}}.
=\dfrac{x^{4}}{4}+\dfrac12\ln|x|+\dfrac{1}{2x^{2}}+c

\dfrac{x^{4}}{4}+\dfrac{1}{2}\ln|x|+\dfrac{1}{2x^{2}}+c

Evaluate \displaystyle\int\frac{2x+1}{x^{2}+3}\,dx.

Solution

Split the fraction into two parts.
\int\dfrac{2x+1}{x^{2}+3}\,dx=\int\dfrac{2x}{x^{2}+3}\,dx+\int\dfrac{1}{x^{2}+3}\,dx
In the first part the numerator 2x is the derivative of the denominator, so it gives \ln(x^{2}+3). No absolute value is needed, since x^{2}+3>0.
\int\dfrac{2x}{x^{2}+3}\,dx=\ln(x^{2}+3)
The second part has the form \int\dfrac{dx}{a^{2}+x^{2}}=\dfrac1a\tan^{-1}\dfrac{x}{a} with a=\sqrt3 (Formula 3.9).
\int\dfrac{dx}{(\sqrt3)^{2}+x^{2}}=\dfrac{1}{\sqrt3}\tan^{-1}\left(\dfrac{x}{\sqrt3}\right)
Check the second part by differentiating, since this formula comes later in the book (Section 3.3).
\dfrac{d}{dx}\left[\dfrac{1}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3}\right]=\dfrac{1}{\sqrt3}\cdot\dfrac{1/\sqrt3}{1+x^{2}/3}=\dfrac{1}{3+x^{2}}\ \checkmark
Add the two parts, with one constant.
\int\dfrac{2x+1}{x^{2}+3}\,dx=\ln(x^{2}+3)+\dfrac{1}{\sqrt3}\tan^{-1}\left(\dfrac{x}{\sqrt3}\right)+c

\ln(x^{2}+3)+\dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{x}{\sqrt{3}}\right)+c

Evaluate \displaystyle\int\frac{e^{\tan^{-1}z}}{1+z^{2}}\,dz.

Solution

The power of e is f(z)=\tan^{-1}z, and its derivative f'(z)=\dfrac{1}{1+z^{2}} is the other factor.
\int\dfrac{e^{\tan^{-1}z}}{1+z^{2}}\,dz=\int e^{\tan^{-1}z}\cdot\dfrac{1}{1+z^{2}}\,dz
Use \int e^{f(z)}f'(z)\,dz=e^{f(z)}+c (or substitute u=\tan^{-1}z).
=e^{\tan^{-1}z}+c

e^{\tan^{-1}z}+c

Evaluate \int\left(x^{3/2}+e^{3x}+x^{0}\right)dx.

Solution

x^{0}=1 (for x\neq0), so the last term is the constant 1.
\int\left(x^{3/2}+e^{3x}+x^{0}\right)dx=\int x^{3/2}\,dx+\int e^{3x}\,dx+\int1\,dx
Use the power rule for x^{3/2} and \int e^{ax}dx=\dfrac{e^{ax}}{a} for e^{3x}.
=\dfrac{x^{5/2}}{5/2}+\dfrac{e^{3x}}{3}+x+c=\dfrac25x^{5/2}+\dfrac13e^{3x}+x+c

\dfrac{2}{5}x^{5/2}+\dfrac{1}{3}e^{3x}+x+c

Evaluate \int(3x^{2}+2x)(x^{3}+x^{2}+9)^{5}\,dx.

Solution

Let f(x)=x^{3}+x^{2}+9. Then f'(x)=3x^{2}+2x, which is exactly the other factor.
\int(3x^{2}+2x)(x^{3}+x^{2}+9)^{5}\,dx=\int\big[f(x)\big]^{5}f'(x)\,dx
Use the general power rule with n=5.
=\dfrac{\big[f(x)\big]^{6}}{6}+c=\dfrac{(x^{3}+x^{2}+9)^{6}}{6}+c

\dfrac{(x^{3}+x^{2}+9)^{6}}{6}+c

Evaluate \int(5e^{5x}-x^{-3}+3^{2x})\,dx.

Solution

Integrate term by term.
\int(5e^{5x}-x^{-3}+3^{2x})\,dx=\int5e^{5x}\,dx-\int x^{-3}\,dx+\int3^{2x}\,dx
5 is the derivative of 5x, so \int5e^{5x}dx=e^{5x}. By the power rule \int x^{-3}dx=\dfrac{x^{-2}}{-2}.
\int5e^{5x}\,dx=e^{5x},\qquad-\int x^{-3}\,dx=-\dfrac{x^{-2}}{-2}=\dfrac{1}{2x^{2}}
For 3^{2x} use \int a^{f(x)}f'(x)\,dx=\dfrac{a^{f(x)}}{\ln a} with f(x)=2x and f'(x)=2: multiply and divide by 2.
\int3^{2x}\,dx=\dfrac12\int3^{2x}(2)\,dx=\dfrac{3^{2x}}{2\ln3}
Add the results.
\int(5e^{5x}-x^{-3}+3^{2x})\,dx=e^{5x}+\dfrac{1}{2x^{2}}+\dfrac{3^{2x}}{2\ln3}+c

e^{5x}+\dfrac{1}{2x^{2}}+\dfrac{3^{2x}}{2\ln 3}+c

Evaluate \displaystyle\int\left(z^{-1/4}+\sqrt{3z}+\frac{4}{z}-\frac{1}{e^{z}}\right)dz.

Solution

Rewrite every term in a standard form: \sqrt{3z}=\sqrt3\,z^{1/2} and \dfrac{1}{e^{z}}=e^{-z}.
z^{-1/4}+\sqrt{3z}+\dfrac4z-\dfrac{1}{e^{z}}=z^{-1/4}+\sqrt3\,z^{1/2}+4\cdot\dfrac1z-e^{-z}
Integrate each term: the power rule for the powers, \int\dfrac{dz}{z}=\ln|z|, and \int e^{-z}dz=-e^{-z}.
=\dfrac{z^{3/4}}{3/4}+\sqrt3\cdot\dfrac{z^{3/2}}{3/2}+4\ln|z|-\left(-e^{-z}\right)+c
Simplify, using \sqrt3\cdot\dfrac23=\dfrac{2\sqrt3}{3}.
=\dfrac43z^{3/4}+\dfrac{2\sqrt3}{3}z^{3/2}+4\ln|z|+e^{-z}+c

\dfrac{4}{3}z^{3/4}+\dfrac{2\sqrt{3}}{3}z^{3/2}+4\ln|z|+e^{-z}+c

Exercise 3.2p. 96

Questions 1 to 10: evaluate the integrals and recheck your answer by differentiating. Questions 11 to 18: evaluate the integral. Use the table of trigonometric integrals (divide by k when the angle is kx) and the identities \sin^{2}A=\tfrac12(1-\cos2A), \cos^{2}A=\tfrac12(1+\cos2A), \tan^{2}A=\sec^{2}A-1, \cot^{2}A=\operatorname{cosec}^{2}A-1.

Evaluate \int(\sin\pi x-3\sin3x)\,dx and recheck your answer by differentiating.

Solution

Use \int\sin kx\,dx=-\dfrac{\cos kx}{k} for each term (k=\pi and k=3).
\int\sin\pi x\,dx=-\dfrac{\cos\pi x}{\pi},\qquad\int3\sin3x\,dx=3\left(-\dfrac{\cos3x}{3}\right)=-\cos3x
Subtract the second result from the first.
\int(\sin\pi x-3\sin3x)\,dx=-\dfrac{\cos\pi x}{\pi}+\cos3x+c
Check by differentiating.
\dfrac{d}{dx}\left[-\dfrac{\cos\pi x}{\pi}+\cos3x\right]=-\dfrac{-\pi\sin\pi x}{\pi}-3\sin3x=\sin\pi x-3\sin3x\ \checkmark

-\dfrac{1}{\pi}\cos\pi x+\cos 3x+c

Evaluate \displaystyle\int-\sec^{2}\left(\frac{3}{2}y\right)dy and recheck your answer by differentiating.

Solution

Take the minus sign out and use \int\sec^{2}ky\,dy=\dfrac{\tan ky}{k} with k=\tfrac32.
\int-\sec^{2}\left(\dfrac32y\right)dy=-\dfrac{\tan\left(\frac32y\right)}{3/2}+c=-\dfrac23\tan\left(\dfrac32y\right)+c
Check by differentiating.
\dfrac{d}{dy}\left[-\dfrac23\tan\left(\dfrac32y\right)\right]=-\dfrac23\cdot\dfrac32\sec^{2}\left(\dfrac32y\right)=-\sec^{2}\left(\dfrac32y\right)\ \checkmark

-\dfrac{2}{3}\tan\left(\dfrac{3}{2}y\right)+c

Evaluate \int[1-8\operatorname{cosec}^{2}(2x)]\,dx and recheck your answer by differentiating.

Solution

Integrate term by term. Use \int\operatorname{cosec}^{2}kx\,dx=-\dfrac{\cot kx}{k} with k=2.
\int\big[1-8\operatorname{cosec}^{2}2x\big]\,dx=x-8\left(-\dfrac{\cot2x}{2}\right)+c=x+4\cot2x+c
Check by differentiating.
\dfrac{d}{dx}\big[x+4\cot2x\big]=1+4\left(-\operatorname{cosec}^{2}2x\right)(2)=1-8\operatorname{cosec}^{2}2x\ \checkmark

x+4\cot 2x+c

Evaluate \int\frac{1}{2}(\operatorname{cosec}^{2}x-\operatorname{cosec} x\cot x)\,dx and recheck your answer by differentiating.

Solution

Take out \tfrac12. Use \int\operatorname{cosec}^{2}x\,dx=-\cot x and \int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x.
\dfrac12\int(\operatorname{cosec}^{2}x-\operatorname{cosec} x\cot x)\,dx=\dfrac12\big[-\cot x-(-\operatorname{cosec} x)\big]+c=\dfrac12(\operatorname{cosec} x-\cot x)+c
Check by differentiating.
\dfrac{d}{dx}\left[\dfrac12(\operatorname{cosec} x-\cot x)\right]=\dfrac12\big(-\operatorname{cosec} x\cot x+\operatorname{cosec}^{2}x\big)=\dfrac12(\operatorname{cosec}^{2}x-\operatorname{cosec} x\cot x)\ \checkmark

\dfrac{1}{2}(\operatorname{cosec} x-\cot x)+c

Evaluate \displaystyle\int\frac{\cos^{2}z}{7}\,dz and recheck your answer by differentiating.

Solution

Take out \tfrac17 and use \cos^{2}z=\dfrac{1+\cos2z}{2}.
\int\dfrac{\cos^{2}z}{7}\,dz=\dfrac17\int\dfrac{1+\cos2z}{2}\,dz=\dfrac{1}{14}\int(1+\cos2z)\,dz
Integrate.
=\dfrac{1}{14}\left(z+\dfrac{\sin2z}{2}\right)+c=\dfrac{z}{14}+\dfrac{\sin2z}{28}+c
Check by differentiating.
\dfrac{d}{dz}\left[\dfrac{z}{14}+\dfrac{\sin2z}{28}\right]=\dfrac{1}{14}+\dfrac{2\cos2z}{28}=\dfrac{1+\cos2z}{14}=\dfrac{2\cos^{2}z}{14}=\dfrac{\cos^{2}z}{7}\ \checkmark

\dfrac{z}{14}+\dfrac{\sin 2z}{28}+c

Evaluate \int(1+\tan^{2}\theta)\,d\theta and recheck your answer by differentiating.

Solution

Use the identity 1+\tan^{2}\theta=\sec^{2}\theta.
\int(1+\tan^{2}\theta)\,d\theta=\int\sec^{2}\theta\,d\theta=\tan\theta+c
Check by differentiating.
\dfrac{d}{d\theta}(\tan\theta)=\sec^{2}\theta=1+\tan^{2}\theta\ \checkmark

\tan\theta+c

Evaluate \displaystyle\int\frac{1+\cos4t}{2}\,dt and recheck your answer by differentiating.

Solution

Split the fraction and integrate each term; \int\cos4t\,dt=\dfrac{\sin4t}{4}.
\int\dfrac{1+\cos4t}{2}\,dt=\dfrac12\int dt+\dfrac12\int\cos4t\,dt=\dfrac{t}{2}+\dfrac12\cdot\dfrac{\sin4t}{4}+c=\dfrac{t}{2}+\dfrac{\sin4t}{8}+c
Check by differentiating.
\dfrac{d}{dt}\left[\dfrac t2+\dfrac{\sin4t}{8}\right]=\dfrac12+\dfrac{4\cos4t}{8}=\dfrac{1+\cos4t}{2}\ \checkmark

\dfrac{t}{2}+\dfrac{\sin 4t}{8}+c

Evaluate \int\sec^{2}(5x-1)\,dx and recheck your answer by differentiating.

Solution

The angle 5x-1 is linear with coefficient 5, so integrate as usual and divide by 5.
\int\sec^{2}(5x-1)\,dx=\dfrac15\int\sec^{2}(5x-1)\,(5)\,dx=\dfrac15\tan(5x-1)+c
Check by differentiating.
\dfrac{d}{dx}\left[\dfrac15\tan(5x-1)\right]=\dfrac15\sec^{2}(5x-1)\cdot5=\sec^{2}(5x-1)\ \checkmark

\dfrac{1}{5}\tan(5x-1)+c

Evaluate \int(\tan5x+\cos7x)\,dx and recheck your answer by differentiating.

Solution

Use \int\tan kx\,dx=\dfrac1k\ln|\sec kx| with k=5, and \int\cos kx\,dx=\dfrac{\sin kx}{k} with k=7.
\int(\tan5x+\cos7x)\,dx=\dfrac15\ln|\sec5x|+\dfrac17\sin7x+c
Check by differentiating.
\dfrac{d}{dx}\left[\dfrac15\ln|\sec5x|+\dfrac17\sin7x\right]=\dfrac15\cdot\dfrac{5\sec5x\tan5x}{\sec5x}+\dfrac{7\cos7x}{7}=\tan5x+\cos7x\ \checkmark

\dfrac{1}{5}\ln|\sec 5x|+\dfrac{1}{7}\sin 7x+c

Evaluate \int(\cot9y-3)\,dy and recheck your answer by differentiating.

Solution

Use \int\cot ky\,dy=\dfrac1k\ln|\sin ky| with k=9, and \int3\,dy=3y.
\int(\cot9y-3)\,dy=\dfrac19\ln|\sin9y|-3y+c
Check by differentiating.
\dfrac{d}{dy}\left[\dfrac19\ln|\sin9y|-3y\right]=\dfrac19\cdot\dfrac{9\cos9y}{\sin9y}-3=\cot9y-3\ \checkmark

\dfrac{1}{9}\ln|\sin 9y|-3y+c

Evaluate \int(\tan^{2}2\theta+\cot^{2}2\theta)\,d\theta.

Solution

Use \tan^{2}A=\sec^{2}A-1 and \cot^{2}A=\operatorname{cosec}^{2}A-1 with A=2\theta.
\tan^{2}2\theta+\cot^{2}2\theta=\sec^{2}2\theta+\operatorname{cosec}^{2}2\theta-2
Integrate each term, dividing by 2 because the angle is 2\theta.
\int(\sec^{2}2\theta+\operatorname{cosec}^{2}2\theta-2)\,d\theta=\dfrac{\tan2\theta}{2}-\dfrac{\cot2\theta}{2}-2\theta+c
Check by differentiating.
\dfrac{d}{d\theta}\left[\dfrac{\tan2\theta-\cot2\theta}{2}-2\theta\right]=\sec^{2}2\theta+\operatorname{cosec}^{2}2\theta-2=\tan^{2}2\theta+\cot^{2}2\theta\ \checkmark

\dfrac{1}{2}\tan 2\theta-\dfrac{1}{2}\cot 2\theta-2\theta+c

Evaluate \displaystyle\int\sin^{2}\left(\frac{11}{2}y\right)dy.

Solution

Use \sin^{2}A=\dfrac{1-\cos2A}{2} with A=\tfrac{11}{2}y, so that 2A=11y.
\int\sin^{2}\left(\dfrac{11}{2}y\right)dy=\int\dfrac{1-\cos11y}{2}\,dy
Integrate; \int\cos11y\,dy=\dfrac{\sin11y}{11}.
=\dfrac{y}{2}-\dfrac12\cdot\dfrac{\sin11y}{11}+c=\dfrac{y}{2}-\dfrac{\sin11y}{22}+c
Check by differentiating.
\dfrac{d}{dy}\left[\dfrac y2-\dfrac{\sin11y}{22}\right]=\dfrac12-\dfrac{\cos11y}{2}=\dfrac{1-\cos11y}{2}=\sin^{2}\left(\dfrac{11}{2}y\right)\ \checkmark

\dfrac{y}{2}-\dfrac{\sin 11y}{22}+c

Evaluate \int\operatorname{cosec}11x\tan11x\,dx.

Solution

Simplify first: \operatorname{cosec}11x\tan11x=\dfrac{1}{\sin11x}\cdot\dfrac{\sin11x}{\cos11x}=\dfrac{1}{\cos11x}=\sec11x.
\int\operatorname{cosec}11x\tan11x\,dx=\int\sec11x\,dx
Use \int\sec u\,du=\ln|\sec u+\tan u| and divide by 11, the coefficient of x.
=\dfrac{1}{11}\ln|\sec11x+\tan11x|+c
Check by differentiating.
\dfrac{d}{dx}\left[\dfrac{1}{11}\ln|\sec11x+\tan11x|\right]=\dfrac{1}{11}\cdot\dfrac{11(\sec11x\tan11x+\sec^{2}11x)}{\sec11x+\tan11x}=\sec11x\ \checkmark

\dfrac{1}{11}\ln|\sec 11x+\tan 11x|+c

Evaluate \int\cos\theta(\tan\theta+\sec\theta)\,d\theta.

Solution

Multiply out: \cos\theta\tan\theta=\cos\theta\cdot\dfrac{\sin\theta}{\cos\theta}=\sin\theta and \cos\theta\sec\theta=1.
\int\cos\theta(\tan\theta+\sec\theta)\,d\theta=\int(\sin\theta+1)\,d\theta
Integrate.
=-\cos\theta+\theta+c
Check by differentiating.
\dfrac{d}{d\theta}(\theta-\cos\theta)=1+\sin\theta=\cos\theta(\tan\theta+\sec\theta)\ \checkmark

\theta-\cos\theta+c

Evaluate \displaystyle\int\operatorname{cosec}^{2}\left(\frac{x-1}{3}\right)dx.

Solution

The angle \dfrac{x-1}{3}=\tfrac13x-\tfrac13 is linear with coefficient \tfrac13. Use \int\operatorname{cosec}^{2}u\,du=-\cot u and divide by \tfrac13, that is, multiply by 3.
\int\operatorname{cosec}^{2}\left(\dfrac{x-1}{3}\right)dx=-\dfrac{\cot\left(\frac{x-1}{3}\right)}{1/3}+c=-3\cot\left(\dfrac{x-1}{3}\right)+c
Check by differentiating.
\dfrac{d}{dx}\left[-3\cot\left(\dfrac{x-1}{3}\right)\right]=-3\left(-\operatorname{cosec}^{2}\dfrac{x-1}{3}\right)\cdot\dfrac13=\operatorname{cosec}^{2}\left(\dfrac{x-1}{3}\right)\ \checkmark

-3\cot\left(\dfrac{x-1}{3}\right)+c

Evaluate \int(\cos x)^{1/5}\sin x\,dx.

Solution

Let f(x)=\cos x, so f'(x)=-\sin x. Put a minus sign inside and outside so that the derivative appears.
\int(\cos x)^{1/5}\sin x\,dx=-\int(\cos x)^{1/5}(-\sin x)\,dx=-\int\big[f(x)\big]^{1/5}f'(x)\,dx
Use the general power rule with n=\tfrac15, so n+1=\tfrac65.
=-\dfrac{(\cos x)^{6/5}}{6/5}+c=-\dfrac56(\cos x)^{6/5}+c
Check by differentiating.
\dfrac{d}{dx}\left[-\dfrac56(\cos x)^{6/5}\right]=-\dfrac56\cdot\dfrac65(\cos x)^{1/5}(-\sin x)=(\cos x)^{1/5}\sin x\ \checkmark

-\dfrac{5}{6}(\cos x)^{6/5}+c

Evaluate \int e^{y}\sin e^{y}\,dy.

Solution

Substitute t=e^{y}. Then dt=e^{y}\,dy, which is exactly the rest of the integrand.
\int\sin(e^{y})\,e^{y}\,dy=\int\sin t\,dt=-\cos t+c
Substitute back t=e^{y}.
=-\cos(e^{y})+c
Check by differentiating.
\dfrac{d}{dy}\big[-\cos(e^{y})\big]=\sin(e^{y})\cdot e^{y}\ \checkmark

-\cos(e^{y})+c

Evaluate \int9\tan(x+7)\,dx.

Solution

Take out 9. The angle x+7 has coefficient 1, so \int\tan(x+7)\,dx=\ln|\sec(x+7)|.
\int9\tan(x+7)\,dx=9\ln|\sec(x+7)|+c
An equivalent form, since \ln|\sec u|=-\ln|\cos u|.
=-9\ln|\cos(x+7)|+c
Check by differentiating.
\dfrac{d}{dx}\big[9\ln|\sec(x+7)|\big]=9\cdot\dfrac{\sec(x+7)\tan(x+7)}{\sec(x+7)}=9\tan(x+7)\ \checkmark

9\ln|\sec(x+7)|+c

Exercise 3.3p. 99

Method: integration by substitution (section 3.3). Look for a part of the integrand whose derivative is also present, apart from a constant factor. Call that part u and replace dx as well. When a quadratic sits under a root or in a denominator, complete the square first and then use Formulas 3.5 to 3.9.

Evaluate \displaystyle\int\frac{dx}{x^2+9}.

Solution

Method: substitution x=3\tan\theta

The denominator has the form x^2+a^2 with a^2=9, so a=3. The substitution x=a\tan\theta turns x^2+a^2 into a^2\sec^2\theta, because 1+\tan^2\theta=\sec^2\theta.
x=3\tan\theta\quad\Rightarrow\quad dx=3\sec^2\theta\,d\theta
Rewrite the denominator in terms of \theta.
x^2+9=9\tan^2\theta+9=9(\tan^2\theta+1)=9\sec^2\theta
Substitute both results. The \sec^2\theta cancels, leaving a constant to integrate.
\int\frac{dx}{x^2+9}=\int\frac{3\sec^2\theta\,d\theta}{9\sec^2\theta}=\frac13\int d\theta=\frac13\,\theta+c
Return to x. From x=3\tan\theta we get \tan\theta=\dfrac{x}{3}, so \theta=\tan^{-1}\dfrac{x}{3}.
\int\frac{dx}{x^2+9}=\frac13\tan^{-1}\left(\frac{x}{3}\right)+c
Check. Differentiate the answer; we should get the integrand back.
\frac{d}{dx}\left[\frac13\tan^{-1}\frac{x}{3}\right]=\frac13\cdot\frac{1}{1+\frac{x^2}{9}}\cdot\frac13=\frac{1}{9+x^2}\ \checkmark

\int\dfrac{dx}{x^2+9}=\dfrac{1}{3}\tan^{-1}\left(\dfrac{x}{3}\right)+c

Evaluate \displaystyle\int\frac{dx}{\sqrt{5-x^2}}.

Solution

Method: substitution x=\sqrt5\sin\theta

The root has the form \sqrt{a^2-x^2} with a^2=5, so a=\sqrt5. Put x=\sqrt5\sin\theta, because 1-\sin^2\theta=\cos^2\theta removes the root.
x=\sqrt5\sin\theta\quad\Rightarrow\quad dx=\sqrt5\cos\theta\,d\theta
Simplify the root.
\sqrt{5-x^2}=\sqrt{5-5\sin^2\theta}=\sqrt5\sqrt{1-\sin^2\theta}=\sqrt5\cos\theta
Substitute. Everything cancels except d\theta.
\int\frac{dx}{\sqrt{5-x^2}}=\int\frac{\sqrt5\cos\theta\,d\theta}{\sqrt5\cos\theta}=\int d\theta=\theta+c
Return to x: \sin\theta=\dfrac{x}{\sqrt5}, so \theta=\sin^{-1}\dfrac{x}{\sqrt5}.
\int\frac{dx}{\sqrt{5-x^2}}=\sin^{-1}\left(\frac{x}{\sqrt5}\right)+c
Check. Differentiate with the chain rule.
\frac{d}{dx}\sin^{-1}\frac{x}{\sqrt5}=\frac{1}{\sqrt{1-\frac{x^2}{5}}}\cdot\frac{1}{\sqrt5}=\frac{1}{\sqrt{5-x^2}}\ \checkmark

\int\dfrac{dx}{\sqrt{5-x^2}}=\sin^{-1}\left(\dfrac{x}{\sqrt{5}}\right)+c

Evaluate \displaystyle\int(2x+7)(x^2+7x+3)^{\frac45}\,dx.

Solution

Method: substitution u=x^2+7x+3

The derivative of the bracket x^2+7x+3 is 2x+7, and 2x+7 is the other factor. So put the bracket equal to u.
u=x^2+7x+3\quad\Rightarrow\quad du=(2x+7)\,dx
Substitute. The whole of (2x+7)\,dx becomes du.
\int(2x+7)(x^2+7x+3)^{\frac45}dx=\int u^{\frac45}\,du
Use the power rule \int u^n\,du=\dfrac{u^{n+1}}{n+1}+c with n=\tfrac45, so n+1=\tfrac95.
=\frac{u^{\frac95}}{\frac95}+c=\frac59\,u^{\frac95}+c
Put back u=x^2+7x+3.
\int(2x+7)(x^2+7x+3)^{\frac45}dx=\frac59(x^2+7x+3)^{\frac95}+c
Check. \dfrac{d}{dx}\left[\dfrac59u^{\frac95}\right]=\dfrac59\cdot\dfrac95u^{\frac45}\cdot\dfrac{du}{dx}.
=(x^2+7x+3)^{\frac45}(2x+7)\ \checkmark

\int(2x+7)(x^2+7x+3)^{\frac{4}{5}}dx=\dfrac{5}{9}(x^2+7x+3)^{\frac{9}{5}}+c

Evaluate \displaystyle\int\frac{x^2}{x^3+1}\,dx.

Solution

Method: substitution u=x^3+1

The derivative of the denominator x^3+1 is 3x^2, and the numerator is x^2 (the same apart from the factor 3). So put u=x^3+1.
u=x^3+1\quad\Rightarrow\quad du=3x^2\,dx\quad\Rightarrow\quad x^2\,dx=\frac{du}{3}
Substitute.
\int\frac{x^2}{x^3+1}\,dx=\int\frac1u\cdot\frac{du}{3}=\frac13\int\frac{du}{u}
Use \int\dfrac{du}{u}=\ln|u|+c and put back u=x^3+1.
=\frac13\ln|u|+c=\frac13\ln|x^3+1|+c

\int\dfrac{x^2}{x^3+1}dx=\dfrac{1}{3}\ln|x^3+1|+c

Evaluate \displaystyle\int\frac{dy}{y^2+8y+20}.

Solution

Method: complete the square, then substitution u=y+4

The denominator does not factorise (b^2-4ac=64-80<0). Complete the square: add and subtract \left(\tfrac82\right)^2=16.
y^2+8y+20=(y^2+8y+16)+4=(y+4)^2+2^2
Put u=y+4. Then du=dy.
\int\frac{dy}{y^2+8y+20}=\int\frac{du}{u^2+2^2}
Use Formula 3.9, \int\dfrac{du}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac{u}{a}+c, with a=2.
=\frac12\tan^{-1}\left(\frac{u}{2}\right)+c
Put back u=y+4.
\int\frac{dy}{y^2+8y+20}=\frac12\tan^{-1}\left(\frac{y+4}{2}\right)+c

\int\dfrac{dy}{y^2+8y+20}=\dfrac{1}{2}\tan^{-1}\left(\dfrac{y+4}{2}\right)+c

Evaluate \displaystyle\int\frac{dx}{\sqrt{20-x^2-4x}}.

Solution

Method: complete the square, then Formula 3.5

Write the quadratic as a number minus a perfect square. Take a minus sign out of the x terms, then complete the square inside the bracket by adding and subtracting 4.
20-x^2-4x=20-(x^2+4x)=20-(x^2+4x+4)+4=24-(x+2)^2
So a^2=24 and a=\sqrt{24}=2\sqrt6. Put u=x+2, so du=dx.
\int\frac{dx}{\sqrt{20-x^2-4x}}=\int\frac{du}{\sqrt{(2\sqrt6)^2-u^2}}
Use Formula 3.5, \int\dfrac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\dfrac{u}{a}+c.
=\sin^{-1}\left(\frac{u}{2\sqrt6}\right)+c
Put back u=x+2.
\int\frac{dx}{\sqrt{20-x^2-4x}}=\sin^{-1}\left(\frac{x+2}{2\sqrt6}\right)+c
Check. Differentiate with the chain rule.
\frac{1}{\sqrt{1-\frac{(x+2)^2}{24}}}\cdot\frac{1}{2\sqrt6}=\frac{1}{\sqrt{24-(x+2)^2}}=\frac{1}{\sqrt{20-x^2-4x}}\ \checkmark

\int\dfrac{dx}{\sqrt{20-x^2-4x}}=\sin^{-1}\left(\dfrac{x+2}{2\sqrt{6}}\right)+c

Evaluate \displaystyle\int\frac{x\,dx}{(4x^2+1)^3}.

Solution

Method: substitution u=4x^2+1

The derivative of 4x^2+1 is 8x, and x\,dx is in the numerator. Put u=4x^2+1.
u=4x^2+1\quad\Rightarrow\quad du=8x\,dx\quad\Rightarrow\quad x\,dx=\frac{du}{8}
Substitute and write the power in the denominator as a negative index.
\int\frac{x\,dx}{(4x^2+1)^3}=\frac18\int\frac{du}{u^3}=\frac18\int u^{-3}\,du
Power rule with n=-3, so n+1=-2.
=\frac18\cdot\frac{u^{-2}}{-2}+c=-\frac{1}{16u^2}+c
Put back u=4x^2+1.
\int\frac{x\,dx}{(4x^2+1)^3}=-\frac{1}{16(4x^2+1)^2}+c

\int\dfrac{x\,dx}{(4x^2+1)^3}=-\dfrac{1}{16(4x^2+1)^2}+c

Evaluate \displaystyle\int x^4\sqrt{3x^5-5}\,dx.

Solution

Method: substitution u=3x^5-5

The derivative of 3x^5-5 is 15x^4, and x^4 stands outside the root. Put u=3x^5-5.
u=3x^5-5\quad\Rightarrow\quad du=15x^4\,dx\quad\Rightarrow\quad x^4\,dx=\frac{du}{15}
Substitute and write the root as the power \tfrac12.
\int x^4\sqrt{3x^5-5}\,dx=\frac1{15}\int u^{\frac12}\,du
Power rule with n=\tfrac12, so n+1=\tfrac32.
=\frac1{15}\cdot\frac{u^{\frac32}}{\frac32}+c=\frac{2}{45}\,u^{\frac32}+c
Put back u=3x^5-5.
\int x^4\sqrt{3x^5-5}\,dx=\frac{2}{45}(3x^5-5)^{\frac32}+c
Check. \dfrac{d}{dx}\left[\dfrac{2}{45}u^{\frac32}\right]=\dfrac{2}{45}\cdot\dfrac32u^{\frac12}\cdot15x^4.
=x^4\sqrt{3x^5-5}\ \checkmark

\int x^4\sqrt{3x^5-5}\,dx=\dfrac{2}{45}(3x^5-5)^{\frac{3}{2}}+c

Evaluate \displaystyle\int\frac{2ax+b}{ax^2+bx+c}\,dx.

Solution

Method: substitution u=ax^2+bx+c

The numerator 2ax+b is exactly the derivative of the denominator. Put u=ax^2+bx+c.
u=ax^2+bx+c\quad\Rightarrow\quad du=(2ax+b)\,dx
Substitute.
\int\frac{2ax+b}{ax^2+bx+c}\,dx=\int\frac{du}{u}=\ln|u|+C
Put back u. We write the constant of integration as C because the letter c is already used in the question.
\int\frac{2ax+b}{ax^2+bx+c}\,dx=\ln|ax^2+bx+c|+C

\int\dfrac{2ax+b}{ax^2+bx+c}dx=\ln|ax^2+bx+c|+C

Evaluate \displaystyle\int\frac{dx}{(1-3x)^2}.

Solution

Method: substitution u=1-3x

Put u=1-3x, the expression inside the bracket.
u=1-3x\quad\Rightarrow\quad du=-3\,dx\quad\Rightarrow\quad dx=-\frac{du}{3}
Substitute.
\int\frac{dx}{(1-3x)^2}=-\frac13\int u^{-2}\,du
Power rule with n=-2, so n+1=-1. The two minus signs cancel.
=-\frac13\cdot\frac{u^{-1}}{-1}+c=\frac{1}{3u}+c
Put back u=1-3x.
\int\frac{dx}{(1-3x)^2}=\frac{1}{3(1-3x)}+c
Check. \dfrac{d}{dx}\left[\dfrac13(1-3x)^{-1}\right]=\dfrac13(-1)(1-3x)^{-2}(-3).
=\frac{1}{(1-3x)^2}\ \checkmark

\int\dfrac{dx}{(1-3x)^2}=\dfrac{1}{3(1-3x)}+c

Evaluate \displaystyle\int\frac{z^3}{1+z^4}\,dz.

Solution

Method: substitution u=1+z^4

The derivative of 1+z^4 is 4z^3, and z^3 is the numerator. Put u=1+z^4.
u=1+z^4\quad\Rightarrow\quad du=4z^3\,dz\quad\Rightarrow\quad z^3\,dz=\frac{du}{4}
Substitute.
\int\frac{z^3}{1+z^4}\,dz=\frac14\int\frac{du}{u}=\frac14\ln u+c
Put back u=1+z^4. No modulus is needed because 1+z^4>0 for every z.
\int\frac{z^3}{1+z^4}\,dz=\frac14\ln(1+z^4)+c

\int\dfrac{z^3}{1+z^4}dz=\dfrac{1}{4}\ln(1+z^4)+c

Evaluate \displaystyle\int\frac{\cot^{-1}x}{1+x^2}\,dx.

Solution

Method: substitution u=\cot^{-1}x

The derivative of \cot^{-1}x is -\dfrac{1}{1+x^2}, and \dfrac{1}{1+x^2} is present. Put u=\cot^{-1}x.
u=\cot^{-1}x\quad\Rightarrow\quad du=-\frac{dx}{1+x^2}\quad\Rightarrow\quad\frac{dx}{1+x^2}=-du
Substitute.
\int\frac{\cot^{-1}x}{1+x^2}\,dx=\int u\,(-du)=-\int u\,du
Integrate with the power rule and put back u=\cot^{-1}x.
=-\frac{u^2}{2}+c=-\frac{(\cot^{-1}x)^2}{2}+c

\int\dfrac{\cot^{-1}x}{1+x^2}dx=-\dfrac{(\cot^{-1}x)^2}{2}+c

Exercise 3.4p. 102

Method: integration by parts (section 3.4), \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int\left[f'(x)\int g(x)\,dx\right]dx. The first function f is differentiated and the second function g is integrated, so choose as f the function that becomes simpler when differentiated. A useful order for choosing f is ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. It agrees with the book's Key Facts. When only one function is given (such as \ln x or \tan^{-1}x), multiply it by 1 and take 1 as the second function.

Evaluate \displaystyle\int\ln x\,dx.

Solution

Method: integration by parts with f=\ln x, g=1

There is only one function, and we cannot integrate \ln x directly. Write it as a product with 1: take \ln x as the first function (we know its derivative) and 1 as the second function.
\int\ln x\,dx=\int(\ln x)(1)\,dx,\qquad\frac{d}{dx}\ln x=\frac1x,\qquad\int1\,dx=x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int(\ln x)(1)\,dx=\ln x\int1\,dx-\int\left[\frac{d}{dx}(\ln x)\int1\,dx\right]dx=x\ln x-\int\frac1x\cdot x\,dx
Simplify \dfrac1x\cdot x=1 and integrate.
=x\ln x-\int1\,dx=x\ln x-x+c

\int\ln x\,dx=x\ln x-x+c

Evaluate \displaystyle\int(\ln x)^2\,dx.

Solution

Method: integration by parts with f=(\ln x)^2, g=1

Write the integrand as (\ln x)^2\cdot1. Take (\ln x)^2 as the first function and 1 as the second function. Differentiate the first function by the chain rule.
\frac{d}{dx}(\ln x)^2=2\ln x\cdot\frac1x,\qquad\int1\,dx=x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int(\ln x)^2dx=(\ln x)^2\cdot x-\int\frac{2\ln x}{x}\cdot x\,dx=x(\ln x)^2-2\int\ln x\,dx
Use the result of Question 1, \int\ln x\,dx=x\ln x-x.
=x(\ln x)^2-2(x\ln x-x)+c
Expand the bracket.
\int(\ln x)^2dx=x(\ln x)^2-2x\ln x+2x+c

\int(\ln x)^2dx=x(\ln x)^2-2x\ln x+2x+c

Evaluate \displaystyle\int\sin(\ln x)\,dx.

Solution

Method: integration by parts twice (the integral returns)

Let I be the required integral and write it as \sin(\ln x)\cdot1. Take \sin(\ln x) as the first function and 1 as the second function.
I=\int\sin(\ln x)\cdot1\,dx,\qquad\frac{d}{dx}\sin(\ln x)=\frac{\cos(\ln x)}{x}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
I=x\sin(\ln x)-\int\frac{\cos(\ln x)}{x}\cdot x\,dx=x\sin(\ln x)-\int\cos(\ln x)\,dx
Apply integration by parts again to \int\cos(\ln x)\cdot1\,dx, using \dfrac{d}{dx}\cos(\ln x)=-\dfrac{\sin(\ln x)}{x}.
\int\cos(\ln x)\,dx=x\cos(\ln x)-\int\left(-\frac{\sin(\ln x)}{x}\right)x\,dx=x\cos(\ln x)+I
Substitute this into the line for I. The integral I appears on both sides, so bring it to the left (the same idea as Example 16).
I=x\sin(\ln x)-x\cos(\ln x)-I\quad\Rightarrow\quad2I=x\sin(\ln x)-x\cos(\ln x)
Divide by 2 and add the constant of integration.
I=\frac{x}{2}\left[\sin(\ln x)-\cos(\ln x)\right]+c

\int\sin(\ln x)\,dx=\dfrac{x}{2}\left[\sin(\ln x)-\cos(\ln x)\right]+c

Evaluate \displaystyle\int x^3\ln x\,dx.

Solution

Method: integration by parts with f=\ln x, g=x^3

The product has a power of x and a logarithm. By the book's rule, the logarithm is the first function (we cannot integrate it directly) and x^3 is the second function.
f=\ln x,\quad f'=\frac1x;\qquad g=x^3,\quad\int x^3dx=\frac{x^4}{4}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x^3\ln x\,dx=\ln x\cdot\frac{x^4}{4}-\int\frac1x\cdot\frac{x^4}{4}\,dx
Simplify \dfrac1x\cdot\dfrac{x^4}{4}=\dfrac{x^3}{4} and integrate.
=\frac{x^4\ln x}{4}-\frac14\cdot\frac{x^4}{4}+c=\frac{x^4\ln x}{4}-\frac{x^4}{16}+c

\int x^3\ln x\,dx=\dfrac{x^4\ln x}{4}-\dfrac{x^4}{16}+c

Evaluate \displaystyle\int y\sin2y\,dy.

Solution

Method: integration by parts with f=y, g=\sin2y

Take the algebraic factor y as the first function, because its derivative is 1. Take \sin2y as the second function.
f=y,\quad f'=1;\qquad\int\sin2y\,dy=-\frac{\cos2y}{2}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int y\sin2y\,dy=y\left(-\frac{\cos2y}{2}\right)-\int1\cdot\left(-\frac{\cos2y}{2}\right)dy
Simplify and integrate \cos2y.
=-\frac{y\cos2y}{2}+\frac12\int\cos2y\,dy=-\frac{y\cos2y}{2}+\frac12\cdot\frac{\sin2y}{2}+c
Tidy up.
\int y\sin2y\,dy=-\frac{y}{2}\cos2y+\frac14\sin2y+c

\int y\sin 2y\,dy=-\dfrac{y}{2}\cos 2y+\dfrac{1}{4}\sin 2y+c

Evaluate \displaystyle\int e^x\cos x\,dx.

Solution

Method: integration by parts twice (the integral returns)

Both factors keep returning when we differentiate or integrate them. As in Example 16, let I be the integral and take \cos x as the first function and e^x as the second function.
I=\int e^x\cos x\,dx=\int(\cos x)(e^x)\,dx
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx. Here \dfrac{d}{dx}\cos x=-\sin x.
I=\cos x\cdot e^x-\int(-\sin x)\,e^x\,dx=e^x\cos x+\int(\sin x)(e^x)\,dx
Apply integration by parts again, keeping the trigonometric function (\sin x) as the first function.
\int(\sin x)(e^x)\,dx=e^x\sin x-\int(\cos x)\,e^x\,dx=e^x\sin x-I
Substitute back. The integral I returns, so collect it on the left.
I=e^x\cos x+e^x\sin x-I\quad\Rightarrow\quad2I=e^x(\sin x+\cos x)
Divide by 2 and add the constant.
I=\frac{e^x}{2}(\sin x+\cos x)+c

\int e^x\cos x\,dx=\dfrac{e^x}{2}(\sin x+\cos x)+c

Evaluate \displaystyle\int x\sec^{-1}x\,dx.

Solution

Method: integration by parts with f=\sec^{-1}x, g=x

Take the inverse trigonometric function as the first function and x as the second function.
f=\sec^{-1}x,\quad f'=\frac{1}{x\sqrt{x^2-1}};\qquad\int x\,dx=\frac{x^2}{2}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x\sec^{-1}x\,dx=\frac{x^2}{2}\sec^{-1}x-\int\frac{1}{x\sqrt{x^2-1}}\cdot\frac{x^2}{2}\,dx=\frac{x^2}{2}\sec^{-1}x-\frac12\int\frac{x}{\sqrt{x^2-1}}\,dx
The last integral has the form \int[f(x)]^nf'(x)\,dx. Put u=x^2-1, so du=2x\,dx.
\int\frac{x\,dx}{\sqrt{x^2-1}}=\frac12\int u^{-\frac12}\,du=\frac12\cdot2u^{\frac12}=\sqrt{x^2-1}
Substitute back.
\int x\sec^{-1}x\,dx=\frac{x^2}{2}\sec^{-1}x-\frac12\sqrt{x^2-1}+c

\int x\sec^{-1}x\,dx=\dfrac{x^2}{2}\sec^{-1}x-\dfrac{1}{2}\sqrt{x^2-1}+c

Note. We use \dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^2-1}}, which holds for x>1 (the usual textbook case). For x<-1 the derivative is \dfrac{1}{|x|\sqrt{x^2-1}} and the sign of the root term changes.

Evaluate \displaystyle\int\ln(2x+3)\,dx.

Solution

Method: integration by parts with f=\ln(2x+3), g=1

Write the integrand as \ln(2x+3)\cdot1. Take \ln(2x+3) as the first function and 1 as the second function.
\frac{d}{dx}\ln(2x+3)=\frac{2}{2x+3},\qquad\int1\,dx=x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int\ln(2x+3)\,dx=x\ln(2x+3)-\int\frac{2x}{2x+3}\,dx
The fraction \dfrac{2x}{2x+3} has the same degree on top and bottom, so divide first: 2x=(2x+3)-3.
\frac{2x}{2x+3}=1-\frac{3}{2x+3}
Integrate term by term. For the second term use \int\dfrac{dx}{2x+3}=\dfrac12\ln(2x+3).
\int\frac{2x}{2x+3}\,dx=x-\frac32\ln(2x+3)
Substitute back and collect the logarithm terms.
\int\ln(2x+3)\,dx=x\ln(2x+3)-x+\frac32\ln(2x+3)+c=\frac{2x+3}{2}\ln(2x+3)-x+c

\int\ln(2x+3)\,dx=\dfrac{2x+3}{2}\ln(2x+3)-x+c

Evaluate \displaystyle\int x^2e^x\,dx.

Solution

Method: integration by parts twice with x^2 first

Take the power of x as the first function, because each differentiation lowers its power. Take e^x as the second function.
f=x^2,\quad f'=2x;\qquad\int e^x\,dx=e^x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x^2e^x\,dx=x^2e^x-\int2x\,e^x\,dx=x^2e^x-2\int xe^x\,dx
Apply integration by parts again to \int xe^x\,dx with x first (this is Example 13).
\int xe^x\,dx=xe^x-\int1\cdot e^x\,dx=xe^x-e^x
Substitute back and take out the common factor e^x.
\int x^2e^x\,dx=x^2e^x-2(xe^x-e^x)+c=e^x(x^2-2x+2)+c

\int x^2e^x\,dx=e^x(x^2-2x+2)+c

Evaluate \displaystyle\int x\cos x\,dx.

Solution

Method: integration by parts with f=x, g=\cos x

Take x as the first function (its derivative is 1) and \cos x as the second function.
f=x,\quad f'=1;\qquad\int\cos x\,dx=\sin x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x\cos x\,dx=x\sin x-\int1\cdot\sin x\,dx
Integrate \sin x: \int\sin x\,dx=-\cos x.
=x\sin x-(-\cos x)+c=x\sin x+\cos x+c

\int x\cos x\,dx=x\sin x+\cos x+c

Evaluate \displaystyle\int\cos^{-1}x\,dx.

Solution

Method: integration by parts with f=\cos^{-1}x, g=1

Write the integrand as \cos^{-1}x\cdot1. The inverse function is the first function and 1 is the second function.
\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}},\qquad\int1\,dx=x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int\cos^{-1}x\,dx=x\cos^{-1}x-\int\left(-\frac{1}{\sqrt{1-x^2}}\right)x\,dx=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}\,dx
For the last integral put u=1-x^2, so du=-2x\,dx and x\,dx=-\dfrac{du}{2}.
\int\frac{x\,dx}{\sqrt{1-x^2}}=-\frac12\int u^{-\frac12}\,du=-u^{\frac12}=-\sqrt{1-x^2}
Substitute back.
\int\cos^{-1}x\,dx=x\cos^{-1}x-\sqrt{1-x^2}+c

\int\cos^{-1}x\,dx=x\cos^{-1}x-\sqrt{1-x^2}+c

Evaluate \displaystyle\int\tan^{-1}x\,dx.

Solution

Method: integration by parts with f=\tan^{-1}x, g=1

Write the integrand as \tan^{-1}x\cdot1, with \tan^{-1}x as the first function.
\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2},\qquad\int1\,dx=x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx
The numerator x is half the derivative of 1+x^2, so this is a logarithm.
\int\frac{x}{1+x^2}\,dx=\frac12\int\frac{2x}{1+x^2}\,dx=\frac12\ln(1+x^2)
Substitute back. No modulus is needed since 1+x^2>0.
\int\tan^{-1}x\,dx=x\tan^{-1}x-\frac12\ln(1+x^2)+c

\int\tan^{-1}x\,dx=x\tan^{-1}x-\dfrac{1}{2}\ln(1+x^2)+c

Evaluate \displaystyle\int x\sec^2x\,dx.

Solution

Method: integration by parts with f=x, g=\sec^2x

Take x as the first function and \sec^2x as the second function, because \int\sec^2x\,dx=\tan x.
f=x,\quad f'=1;\qquad\int\sec^2x\,dx=\tan x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x\sec^2x\,dx=x\tan x-\int\tan x\,dx
Use \int\tan x\,dx=\ln|\sec x|+c from Example 5 (iii).
\int x\sec^2x\,dx=x\tan x-\ln|\sec x|+c

\int x\sec^2x\,dx=x\tan x-\ln|\sec x|+c

Evaluate \displaystyle\int x^2\sin^{-1}x\,dx.

Solution

Method: integration by parts, then substitution u=1-x^2

Take the inverse function \sin^{-1}x as the first function and x^2 as the second function.
\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}},\qquad\int x^2dx=\frac{x^3}{3}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x^2\sin^{-1}x\,dx=\frac{x^3}{3}\sin^{-1}x-\frac13\int\frac{x^3}{\sqrt{1-x^2}}\,dx
For the last integral put u=1-x^2. Then du=-2x\,dx, and write x^3\,dx=x^2\cdot x\,dx with x^2=1-u.
\int\frac{x^2\cdot x\,dx}{\sqrt{1-x^2}}=\int\frac{1-u}{\sqrt u}\left(-\frac{du}{2}\right)=-\frac12\int\left(u^{-\frac12}-u^{\frac12}\right)du
Integrate each power and put back u=1-x^2.
=-\frac12\left(2u^{\frac12}-\frac23u^{\frac32}\right)=-\sqrt{1-x^2}+\frac13(1-x^2)^{\frac32}
Substitute into step 2. Multiplying by -\tfrac13 changes both signs.
\int x^2\sin^{-1}x\,dx=\frac{x^3}{3}\sin^{-1}x+\frac13\sqrt{1-x^2}-\frac19(1-x^2)^{\frac32}+c
Optional tidy form: take out \dfrac19\sqrt{1-x^2}, since 3-(1-x^2)=x^2+2.
=\frac{x^3}{3}\sin^{-1}x+\frac{(x^2+2)\sqrt{1-x^2}}{9}+c

\int x^2\sin^{-1}x\,dx=\dfrac{x^3}{3}\sin^{-1}x+\dfrac{(x^2+2)\sqrt{1-x^2}}{9}+c

Evaluate \displaystyle\int\ln\left[x+\sqrt{1+x^2}\right]dx.

Solution

Method: integration by parts with f=\ln\left(x+\sqrt{1+x^2}\right), g=1

Write the integrand as \ln\left(x+\sqrt{1+x^2}\right)\cdot1 and take the logarithm as the first function. Find its derivative by the chain rule, then simplify.
\frac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right)=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx. Here \int1\,dx=x.
\int\ln\left(x+\sqrt{1+x^2}\right)dx=x\ln\left(x+\sqrt{1+x^2}\right)-\int\frac{x}{\sqrt{1+x^2}}\,dx
Put u=1+x^2, so du=2x\,dx.
\int\frac{x\,dx}{\sqrt{1+x^2}}=\frac12\int u^{-\frac12}\,du=u^{\frac12}=\sqrt{1+x^2}
Substitute back.
\int\ln\left(x+\sqrt{1+x^2}\right)dx=x\ln\left(x+\sqrt{1+x^2}\right)-\sqrt{1+x^2}+c

\int\ln\left(x+\sqrt{1+x^2}\right)dx=x\ln\left(x+\sqrt{1+x^2}\right)-\sqrt{1+x^2}+c

Evaluate \displaystyle\int x^3e^{x^2}\,dx.

Solution

Method: integration by parts with f=x^2, g=xe^{x^2}

Split x^3=x^2\cdot x. We can integrate xe^{x^2} by substitution, so take x^2 as the first function and xe^{x^2} as the second function.
\int xe^{x^2}dx=\frac12\int e^{x^2}(2x)\,dx=\frac12e^{x^2}
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx. Here f=x^2 and f'=2x.
\int x^3e^{x^2}dx=\int(x^2)\left(xe^{x^2}\right)dx=x^2\cdot\frac12e^{x^2}-\int2x\cdot\frac12e^{x^2}\,dx
The remaining integral is the one from step 1.
=\frac12x^2e^{x^2}-\int xe^{x^2}\,dx=\frac12x^2e^{x^2}-\frac12e^{x^2}+c
Take out the common factor.
\int x^3e^{x^2}dx=\frac12e^{x^2}(x^2-1)+c

\int x^3e^{x^2}dx=\dfrac{1}{2}e^{x^2}(x^2-1)+c

Evaluate \displaystyle\int x^2\sin x\,dx.

Solution

Method: integration by parts twice with x^2 first

Take x^2 as the first function (two differentiations remove it) and \sin x as the second function.
f=x^2,\quad f'=2x;\qquad\int\sin x\,dx=-\cos x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int x^2\sin x\,dx=x^2(-\cos x)-\int2x(-\cos x)\,dx=-x^2\cos x+2\int x\cos x\,dx
Use Question 10: \int x\cos x\,dx=x\sin x+\cos x.
=-x^2\cos x+2(x\sin x+\cos x)+c
Expand.
\int x^2\sin x\,dx=-x^2\cos x+2x\sin x+2\cos x+c

\int x^2\sin x\,dx=-x^2\cos x+2x\sin x+2\cos x+c

Evaluate \displaystyle\int\frac{\ln x}{\sqrt x}\,dx.

Solution

Method: integration by parts with f=\ln x, g=x^{-\frac12}

Write the integrand as (\ln x)\left(x^{-\frac12}\right). Take \ln x as the first function and x^{-\frac12} as the second function.
f=\ln x,\quad f'=\frac1x;\qquad\int x^{-\frac12}dx=\frac{x^{\frac12}}{\frac12}=2\sqrt x
Apply the formula for integration by parts, \int f g\,dx=f\int g\,dx-\int\left[f'\int g\,dx\right]dx.
\int\frac{\ln x}{\sqrt x}\,dx=2\sqrt x\,\ln x-\int\frac1x\cdot2\sqrt x\,dx=2\sqrt x\,\ln x-2\int x^{-\frac12}\,dx
Integrate x^{-\frac12} again.
=2\sqrt x\,\ln x-2\left(2\sqrt x\right)+c=2\sqrt x\,\ln x-4\sqrt x+c
Take out the common factor 2\sqrt x.
\int\frac{\ln x}{\sqrt x}\,dx=2\sqrt x\,(\ln x-2)+c

\int\dfrac{\ln x}{\sqrt{x}}dx=2\sqrt{x}(\ln x-2)+c

Exercise 3.5p. 104

Method: integration by partial fractions (section 3.5). First make sure the fraction is proper (degree of the numerator less than degree of the denominator); if it is not, divide first. Factorise the denominator and write one fraction for each factor: \dfrac{A}{x-a} for a linear factor, \dfrac{A}{x-a}+\dfrac{B}{(x-a)^2} for a repeated factor, and \dfrac{Bx+C}{x^2+px+q} for a quadratic factor that does not factorise. Find the constants by putting convenient values of x or by comparing coefficients, then integrate each fraction.

Evaluate \displaystyle\int\frac{3x+7}{(x+2)(x+3)}\,dx.

Solution

Method: partial fractions with two linear factors

The fraction is proper (degree 1 over degree 2) and the denominator has two different linear factors, so write one fraction for each factor.
\frac{3x+7}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}
Multiply both sides by (x+2)(x+3) to clear the denominators.
3x+7=A(x+3)+B(x+2)
Put x=-2 (this removes the B term), then x=-3 (this removes the A term).
x=-2:\ 1=A(1)\Rightarrow A=1;\qquad x=-3:\ -2=B(-1)\Rightarrow B=2
Integrate term by term, using \int\dfrac{dx}{x+k}=\ln|x+k|.
\int\frac{3x+7}{(x+2)(x+3)}\,dx=\int\frac{dx}{x+2}+2\int\frac{dx}{x+3}=\ln|x+2|+2\ln|x+3|+c
Check the fractions. Put x=0 in both sides of step 1: left side \dfrac76, right side \dfrac12+\dfrac23=\dfrac76.
\frac{7}{6}=\frac{7}{6}\ \checkmark

\int\dfrac{3x+7}{(x+2)(x+3)}dx=\ln|x+2|+2\ln|x+3|+c

Evaluate \displaystyle\int\frac{4x+9}{x^2+x-12}\,dx.

Solution

Method: factorise, then partial fractions

Factorise the denominator: two numbers with product -12 and sum 1 are 4 and -3.
x^2+x-12=(x+4)(x-3)
Write the partial fractions and clear the denominators.
\frac{4x+9}{(x+4)(x-3)}=\frac{A}{x+4}+\frac{B}{x-3}\quad\Rightarrow\quad4x+9=A(x-3)+B(x+4)
Put x=3, then x=-4.
x=3:\ 21=7B\Rightarrow B=3;\qquad x=-4:\ -7=-7A\Rightarrow A=1
Integrate term by term.
\int\frac{4x+9}{x^2+x-12}\,dx=\int\frac{dx}{x+4}+3\int\frac{dx}{x-3}=\ln|x+4|+3\ln|x-3|+c

\int\dfrac{4x+9}{x^2+x-12}dx=\ln|x+4|+3\ln|x-3|+c

Evaluate \displaystyle\int\frac{21-8x}{x^2+x-6}\,dx.

Solution

Method: factorise, then partial fractions

Factorise the denominator: numbers with product -6 and sum 1 are 3 and -2.
x^2+x-6=(x+3)(x-2)
Write the partial fractions and clear the denominators.
\frac{21-8x}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}\quad\Rightarrow\quad21-8x=A(x-2)+B(x+3)
Put x=2, then x=-3.
x=2:\ 5=5B\Rightarrow B=1;\qquad x=-3:\ 45=-5A\Rightarrow A=-9
Integrate term by term.
\int\frac{21-8x}{x^2+x-6}\,dx=-9\int\frac{dx}{x+3}+\int\frac{dx}{x-2}=-9\ln|x+3|+\ln|x-2|+c

\int\dfrac{21-8x}{x^2+x-6}dx=-9\ln|x+3|+\ln|x-2|+c

Evaluate \displaystyle\int\frac{3x+7}{(x+2)^2}\,dx.

Solution

Method: partial fractions with a repeated linear factor

The factor (x+2) is repeated, so we need one fraction for each power of it.
\frac{3x+7}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}\quad\Rightarrow\quad3x+7=A(x+2)+B
Put x=-2 to find B. Then compare the coefficients of x to find A.
x=-2:\ 1=B;\qquad\text{coefficient of }x:\ 3=A
Integrate. For the second fraction use the power rule: \int(x+2)^{-2}dx=-(x+2)^{-1}.
\int\frac{3x+7}{(x+2)^2}\,dx=3\int\frac{dx}{x+2}+\int(x+2)^{-2}\,dx=3\ln|x+2|-\frac{1}{x+2}+c

\int\dfrac{3x+7}{(x+2)^2}dx=3\ln|x+2|-\dfrac{1}{x+2}+c

Evaluate \displaystyle\int\frac{5x^2-5x+2}{(x+1)(x-1)^2}\,dx.

Solution

Method: partial fractions with a repeated linear factor

There is a linear factor (x+1) and a repeated factor (x-1)^2, so three fractions are needed.
\frac{5x^2-5x+2}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}
Clear the denominators.
5x^2-5x+2=A(x-1)^2+B(x+1)(x-1)+C(x+1)
Put x=1, then x=-1.
x=1:\ 2=2C\Rightarrow C=1;\qquad x=-1:\ 12=4A\Rightarrow A=3
No value of x removes both A and C, so compare the coefficients of x^2 to find B.
x^2:\ 5=A+B\quad\Rightarrow\quad B=5-3=2
Integrate each fraction; \int(x-1)^{-2}dx=-(x-1)^{-1}.
\int\frac{5x^2-5x+2}{(x+1)(x-1)^2}\,dx=3\ln|x+1|+2\ln|x-1|-\frac{1}{x-1}+c

\int\dfrac{5x^2-5x+2}{(x+1)(x-1)^2}dx=3\ln|x+1|+2\ln|x-1|-\dfrac{1}{x-1}+c

Evaluate \displaystyle\int\frac{9x^2+3x+29}{(x+1)(x^2+4)}\,dx.

Solution

Method: partial fractions with an irreducible quadratic factor

The quadratic x^2+4 cannot be factorised, so its fraction gets a linear numerator Bx+C.
\frac{9x^2+3x+29}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}
Clear the denominators.
9x^2+3x+29=A(x^2+4)+(Bx+C)(x+1)
Put x=-1 to find A.
9-3+29=5A\quad\Rightarrow\quad A=7
Compare the coefficients of x^2 and the constant terms. (Check with x: 3=B+C is true.)
x^2:\ 9=A+B\Rightarrow B=2;\qquad\text{constants}:\ 29=4A+C\Rightarrow C=1
Split the second fraction into a logarithm part (numerator is the derivative of x^2+4) and a \tan^{-1} part (Formula 3.9 with a=2).
\int\frac{2x+1}{x^2+4}\,dx=\int\frac{2x}{x^2+4}\,dx+\int\frac{dx}{x^2+2^2}=\ln(x^2+4)+\frac12\tan^{-1}\frac{x}{2}
Collect all the terms.
\int\frac{9x^2+3x+29}{(x+1)(x^2+4)}\,dx=7\ln|x+1|+\ln(x^2+4)+\frac12\tan^{-1}\left(\frac{x}{2}\right)+c

\int\dfrac{9x^2+3x+29}{(x+1)(x^2+4)}dx=7\ln|x+1|+\ln(x^2+4)+\dfrac{1}{2}\tan^{-1}\left(\dfrac{x}{2}\right)+c

Evaluate \displaystyle\int\frac{7x^2+7x+4}{(2x+1)(x^2+x+1)}\,dx.

Solution

Method: partial fractions with an irreducible quadratic factor

The quadratic x^2+x+1 does not factorise (b^2-4ac=1-4<0), so its fraction gets a linear numerator.
\frac{7x^2+7x+4}{(2x+1)(x^2+x+1)}=\frac{A}{2x+1}+\frac{Bx+C}{x^2+x+1}
Clear the denominators.
7x^2+7x+4=A(x^2+x+1)+(Bx+C)(2x+1)
Put x=-\tfrac12, which makes 2x+1=0.
\frac74-\frac72+4=A\left(\frac14-\frac12+1\right)\quad\Rightarrow\quad\frac94=\frac34A\quad\Rightarrow\quad A=3
Compare the coefficients of x^2 and the constant terms. (Check with x: 7=A+B+2C=3+2+2.)
x^2:\ 7=A+2B\Rightarrow B=2;\qquad\text{constants}:\ 4=A+C\Rightarrow C=1
Integrate. In the second fraction the numerator 2x+1 is the derivative of x^2+x+1; for the first use \int\dfrac{dx}{2x+1}=\dfrac12\ln|2x+1|.
\int\frac{7x^2+7x+4}{(2x+1)(x^2+x+1)}\,dx=3\int\frac{dx}{2x+1}+\int\frac{2x+1}{x^2+x+1}\,dx=\frac32\ln|2x+1|+\ln(x^2+x+1)+c

\int\dfrac{7x^2+7x+4}{(2x+1)(x^2+x+1)}dx=\dfrac{3}{2}\ln|2x+1|+\ln(x^2+x+1)+c

Evaluate \displaystyle\int\frac{x^3+4x^2+9x+14}{x^2+4x+3}\,dx.

Solution

Method: long division, then partial fractions

The numerator has degree 3 and the denominator degree 2, so the fraction is improper. Divide first (as in Example 17): x(x^2+4x+3)=x^3+4x^2+3x, and subtracting this from the numerator leaves 6x+14.
\frac{x^3+4x^2+9x+14}{x^2+4x+3}=x+\frac{6x+14}{x^2+4x+3}
Factorise x^2+4x+3=(x+1)(x+3) and write partial fractions for the proper part.
\frac{6x+14}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}\quad\Rightarrow\quad6x+14=A(x+3)+B(x+1)
Put x=-1, then x=-3.
x=-1:\ 8=2A\Rightarrow A=4;\qquad x=-3:\ -4=-2B\Rightarrow B=2
Integrate term by term.
\int\frac{x^3+4x^2+9x+14}{x^2+4x+3}\,dx=\int x\,dx+4\int\frac{dx}{x+1}+2\int\frac{dx}{x+3}=\frac{x^2}{2}+4\ln|x+1|+2\ln|x+3|+c

\int\dfrac{x^3+4x^2+9x+14}{x^2+4x+3}dx=\dfrac{x^2}{2}+4\ln|x+1|+2\ln|x+3|+c

Evaluate \displaystyle\int\frac{1}{x^2-9}\,dx.

Solution

Method: partial fractions with two linear factors

Factorise the difference of squares.
x^2-9=(x-3)(x+3)
Write the partial fractions and clear the denominators.
\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}\quad\Rightarrow\quad1=A(x+3)+B(x-3)
Put x=3, then x=-3.
x=3:\ 1=6A\Rightarrow A=\frac16;\qquad x=-3:\ 1=-6B\Rightarrow B=-\frac16
Integrate and combine the logarithms with \ln p-\ln q=\ln\dfrac pq.
\int\frac{dx}{x^2-9}=\frac16\ln|x-3|-\frac16\ln|x+3|+c=\frac16\ln\left|\frac{x-3}{x+3}\right|+c

\int\dfrac{dx}{x^2-9}=\dfrac{1}{6}\ln\left|\dfrac{x-3}{x+3}\right|+c

Evaluate \displaystyle\int\frac{1}{x^3+2x^2+x}\,dx.

Solution

Method: factorise, then partial fractions with a repeated factor

Factorise the denominator: take out x, then recognise a perfect square.
x^3+2x^2+x=x(x^2+2x+1)=x(x+1)^2
There is a linear factor x and a repeated factor (x+1)^2.
\begin{aligned}\frac{1}{x(x+1)^2}&=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\\ \Rightarrow\ 1&=A(x+1)^2\\ &\quad+Bx(x+1)+Cx\end{aligned}
Put x=0, then x=-1.
x=0:\ 1=A;\qquad x=-1:\ 1=-C\Rightarrow C=-1
Compare the coefficients of x^2 to find B.
x^2:\ 0=A+B\quad\Rightarrow\quad B=-1
Integrate. Note \int-(x+1)^{-2}dx=+(x+1)^{-1}.
\int\frac{dx}{x^3+2x^2+x}=\ln|x|-\ln|x+1|+\frac{1}{x+1}+c=\ln\left|\frac{x}{x+1}\right|+\frac{1}{x+1}+c

\int\dfrac{dx}{x^3+2x^2+x}=\ln\left|\dfrac{x}{x+1}\right|+\dfrac{1}{x+1}+c

Evaluate \displaystyle\int\frac{e^x}{(e^x+1)^2(e^x-2)}\,dx.

Solution

Method: substitution t=e^x, then partial fractions

Put t=e^x. Then dt=e^x\,dx, which is exactly the numerator, and the integrand becomes a rational function of t.
\int\frac{e^x\,dx}{(e^x+1)^2(e^x-2)}=\int\frac{dt}{(t+1)^2(t-2)}
Write partial fractions in t (a linear factor and a repeated factor) and clear the denominators.
\begin{aligned}\frac{1}{(t+1)^2(t-2)}&=\frac{A}{t-2}+\frac{B}{t+1}\\ &\quad+\frac{C}{(t+1)^2}\\ \Rightarrow\ 1&=A(t+1)^2\\ &\quad+B(t+1)(t-2)\\ &\quad+C(t-2)\end{aligned}
Put t=2, then t=-1.
t=2:\ 1=9A\Rightarrow A=\frac19;\qquad t=-1:\ 1=-3C\Rightarrow C=-\frac13
Compare the coefficients of t^2.
t^2:\ 0=A+B\quad\Rightarrow\quad B=-\frac19
Integrate each fraction; \int-\tfrac13(t+1)^{-2}dt=\tfrac13(t+1)^{-1}.
\int\frac{dt}{(t+1)^2(t-2)}=\frac19\ln|t-2|-\frac19\ln|t+1|+\frac{1}{3(t+1)}+c
Put back t=e^x and combine the logarithms.
\int\frac{e^x\,dx}{(e^x+1)^2(e^x-2)}=\frac19\ln\left|\frac{e^x-2}{e^x+1}\right|+\frac{1}{3(e^x+1)}+c

\int\dfrac{e^x\,dx}{(e^x+1)^2(e^x-2)}=\dfrac{1}{9}\ln\left|\dfrac{e^x-2}{e^x+1}\right|+\dfrac{1}{3(e^x+1)}+c

Evaluate \displaystyle\int\frac{x}{(x+1)^2(x^2+1)}\,dx.

Solution

Method: partial fractions with repeated and quadratic factors

The denominator has a repeated linear factor (x+1)^2 and a quadratic factor x^2+1 that does not factorise.
\frac{x}{(x+1)^2(x^2+1)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+1}
Clear the denominators.
x=A(x+1)(x^2+1)+B(x^2+1)+(Cx+D)(x+1)^2
Put x=-1 to find B.
-1=2B\quad\Rightarrow\quad B=-\frac12
Expand the right side and compare the coefficients of each power of x.
\begin{aligned}x^3&:\ A+C=0\\ x^2&:\ A+B+2C+D=0\\ x^1&:\ A+C+2D=1\\ x^0&:\ A+B+D=0\end{aligned}
Solve. The x^3 and x^1 equations give 2D=1. The x^0 equation then gives A, and the x^3 equation gives C.
D=\frac12,\qquad A=-B-D=\frac12-\frac12=0,\qquad C=-A=0
So only two fractions remain. Integrate them, using \int\dfrac{dx}{x^2+1}=\tan^{-1}x.
\int\frac{x\,dx}{(x+1)^2(x^2+1)}=-\frac12\int\frac{dx}{(x+1)^2}+\frac12\int\frac{dx}{x^2+1}=\frac{1}{2(x+1)}+\frac12\tan^{-1}x+c

\int\dfrac{x\,dx}{(x+1)^2(x^2+1)}=\dfrac{1}{2(x+1)}+\dfrac{1}{2}\tan^{-1}x+c

Exercise 3.6p. 108–109

Method: the definite integral as net signed area (sections 3.6 and 3.7). Sketch the integrand, find the areas with formulas from geometry, and count a region below the x-axis as negative. Use the properties of the definite integral: equal limits give 0, interchanging the limits changes the sign, an interval can be split at any point c, and constants and sums can be taken outside the integral sign.

Sketch the region where area is represented by the definite integral and evaluate the integral using an appropriate formula from geometry.

Solution

Method. Draw the integrand between the limits, recognise the shape (triangle, trapezium or rectangle) and use its area formula. A region above the x-axis counts as positive and a region below the x-axis counts as negative, because the definite integral is the net signed area (Theorem 3.2).

(i) \displaystyle\int_0^4x\,dx

The integrand is the line y=x, through the origin with slope 1. At the limits: x=0\Rightarrow y=0 and x=4\Rightarrow y=4.
Between x=0 and x=4 the line is above the x-axis, so the region is a right triangle with base 4 and height 4.
Use the area of a triangle, \tfrac12\times\text{base}\times\text{height}. The region is above the axis, so the integral is positive.
\int_0^4x\,dx=\frac12(4)(4)=8

The region is a triangle above the x-axis.

\int_0^4x\,dx=8

(ii) \displaystyle\int_{-3}^0x\,dx

The integrand is again y=x. At the limits: x=-3\Rightarrow y=-3 and x=0\Rightarrow y=0.
Between x=-3 and x=0 the line is below the x-axis. The region is a right triangle with base 3 and height 3.
\text{area}=\frac12(3)(3)=\frac92
A region below the x-axis gives a negative integral.
\int_{-3}^0x\,dx=-\frac92=-4.5
Check with the Fundamental Theorem of Calculus (section 3.8).
\left[\frac{x^2}{2}\right]_{-3}^0=0-\frac92=-\frac92\ \checkmark

The triangle lies below the x-axis, so it counts as negative.

\int_{-3}^{0}x\,dx=-\dfrac{9}{2}

(iii) \displaystyle\int_0^2(x-1)\,dx

The integrand is the line y=x-1. It meets the x-axis at x=1. At the limits: x=0\Rightarrow y=-1 and x=2\Rightarrow y=1.
The region is two triangles: A_1 from x=0 to x=1 (below the axis) and A_2 from x=1 to x=2 (above the axis). Each has base 1 and height 1.
A_1=A_2=\frac12(1)(1)=\frac12
Split the integral at x=1 (Property (iii)). The part below the axis counts as negative.
\int_0^2(x-1)\,dx=\int_0^1(x-1)\,dx+\int_1^2(x-1)\,dx=-\frac12+\frac12=0
The total area of the shaded region, with both triangles counted as positive, is \tfrac12+\tfrac12=1 square unit. The integral and the area differ because part of the region is below the x-axis.
Check with the Fundamental Theorem of Calculus (section 3.8).
\left[\frac{x^2}{2}-x\right]_0^2=(2-2)-(0-0)=0\ \checkmark

A_1 is below the x-axis and A_2 is above it, so they cancel in the integral.

\int_0^2(x-1)\,dx=0\quad(\text{total area}=1\text{ sq. unit})

(iv) \displaystyle\int_0^2(x+1)\,dx

The integrand is the line y=x+1. At the limits: x=0\Rightarrow y=1 and x=2\Rightarrow y=3.
The region between the line and the x-axis from x=0 to x=2 is a trapezium. Its parallel sides are the vertical lengths 1 and 3, and its width is 2.
Use the area of a trapezium, \tfrac12(\text{sum of parallel sides})\times\text{width}. The region is above the axis.
\int_0^2(x+1)\,dx=\frac12(1+3)(2)=4

A trapezium with parallel sides 1 and 3 and width 2.

\int_0^2(x+1)\,dx=4

(v) \displaystyle\int_{-3}^3 2\,dx

The integrand is the constant function y=2, a horizontal line 2 units above the x-axis.
From x=-3 to x=3 the region is a rectangle of width 3-(-3)=6 and height 2, above the axis.
\int_{-3}^3 2\,dx=6\times2=12

A rectangle of width 6 and height 2.

\int_{-3}^{3}2\,dx=12

Note. The book's Example 21 writes \int_0^1(x-1)\,dx=\frac12 and \int_0^2(x-1)\,dx=1 sq. unit. Those numbers are areas of the shaded regions, not values of the integrals. Since a definite integral is the net signed area (Theorem 3.2), the correct values are \int_0^1(x-1)\,dx=-\frac12 (the book itself uses this value in its example of Property (ii)) and \int_0^2(x-1)\,dx=0. Parts (ii) and (iii) above follow the correct rule and also state the area.

Evaluate the integrals in each part when

f(x)=\begin{cases}x; & x\le1\\ 3; & x>1\end{cases}

Solution

Method. The rule for f changes at x=1. For x\le1 the graph is the line y=x (filled dot at (1,1)), and for x>1 it is the horizontal line y=3 (hollow dot at (1,3)). Split each integral at x=1 (Property (iii)) and find every piece as the area of a triangle or a rectangle, with a minus sign for any part below the x-axis.

(i) \displaystyle\int_0^1f(x)\,dx

On [0,1] every x satisfies x\le1, so f(x)=x.
The region under y=x from 0 to 1 is a triangle with base 1 and height 1, above the axis.
\int_0^1f(x)\,dx=\int_0^1x\,dx=\frac12(1)(1)=\frac12

\int_0^1f(x)\,dx=\dfrac{1}{2}

(ii) \displaystyle\int_{-1}^1f(x)\,dx

On [-1,1] we have x\le1, so again f(x)=x.
The region is two triangles with base 1 and height 1: from -1 to 0 below the axis (counts -\tfrac12) and from 0 to 1 above the axis (counts +\tfrac12).
\int_{-1}^1f(x)\,dx=\int_{-1}^0x\,dx+\int_0^1x\,dx=-\frac12+\frac12=0

\int_{-1}^{1}f(x)\,dx=0

(iii) \displaystyle\int_1^4f(x)\,dx

For 1<x\le4 we have f(x)=3. The value f(1)=1 at the single point x=1 does not change the area, because a point has no width.
The region under y=3 from 1 to 4 is a rectangle of width 4-1=3 and height 3.
\int_1^4f(x)\,dx=\int_1^4 3\,dx=3\times3=9

\int_1^4f(x)\,dx=9

(iv) \displaystyle\int_{-1}^2f(x)\,dx

The interval [-1,2] contains x=1, where the rule for f changes. Split the integral there (Property (iii)).
\int_{-1}^2f(x)\,dx=\int_{-1}^1x\,dx+\int_1^2 3\,dx
The first integral is 0 (part (ii)). The second is a rectangle of width 1 and height 3.
=0+3(1)=3

\int_{-1}^{2}f(x)\,dx=3

Using the area shown below in the figure, evaluate the integrals.

The graph of y=f(x) encloses A_1=2 above the x-axis from a to b, A_2=3 below it from b to c, and A_3=3.5 above it from c to d.

Solution

Method. A definite integral is the net signed area. A region above the x-axis adds its area and a region below the x-axis subtracts its area. To cross several regions, split the integral at the crossing points and add the pieces (Property (iii)).

Interval	Region	Position	Signed value
[a,b]	A_1=2	above the x-axis	+2
[b,c]	A_2=3	below the x-axis	-3
[c,d]	A_3=3.5	above the x-axis	+3.5

(i) \displaystyle\int_a^bf(x)\,dx

Only region A_1 lies between a and b, and it is above the axis.
\int_a^bf(x)\,dx=A_1=2

\int_a^bf(x)\,dx=2

(ii) \displaystyle\int_b^cf(x)\,dx

Only region A_2 lies between b and c, and it is below the axis, so it counts as negative.
\int_b^cf(x)\,dx=-A_2=-3

\int_b^cf(x)\,dx=-3

(iii) \displaystyle\int_c^df(x)\,dx

Only region A_3 lies between c and d, above the axis.
\int_c^df(x)\,dx=A_3=3.5

\int_c^df(x)\,dx=3.5

(iv) \displaystyle\int_a^cf(x)\,dx

Split at b (Property (iii)) and add the signed areas.
\int_a^cf(x)\,dx=\int_a^bf(x)\,dx+\int_b^cf(x)\,dx=2+(-3)=-1

\int_a^cf(x)\,dx=-1

(v) \displaystyle\int_b^df(x)\,dx

Split at c and add the signed areas.
\int_b^df(x)\,dx=\int_b^cf(x)\,dx+\int_c^df(x)\,dx=-3+3.5=0.5

\int_b^df(x)\,dx=0.5

(vi) \displaystyle\int_a^df(x)\,dx

Split at b and at c, then add all three signed areas.
\int_a^df(x)\,dx=2+(-3)+3.5=2.5

\int_a^df(x)\,dx=2.5

Note. The book's figure labels the third region A_1=3.5. It is the third region, so it is called A_3 here.

Find: \displaystyle\int_1^5[3f(x)-2g(x)]\,dx if \displaystyle\int_1^5f(x)\,dx=4 and \displaystyle\int_1^5g(x)\,dx=5.

Solution

Method: properties of the definite integral (Theorem 3.3)

Split the integral of a difference into a difference of integrals, and take the constants outside (Theorem 3.3 (iii)).
\int_1^5[3f(x)-2g(x)]\,dx=3\int_1^5f(x)\,dx-2\int_1^5g(x)\,dx
Substitute the given values.
=3(4)-2(5)=12-10=2

\int_1^5[3f(x)-2g(x)]\,dx=2

Find: \displaystyle\int_1^4f(x)\,dx if \displaystyle\int_1^2f(x)\,dx=1 and \displaystyle\int_2^4f(x)\,dx=2.

Solution

Method: splitting the interval (Property (iii))

The point 2 lies between 1 and 4, so split the interval there (Property (iii)).
\int_1^4f(x)\,dx=\int_1^2f(x)\,dx+\int_2^4f(x)\,dx
Substitute the given values.
=1+2=3

\int_1^4f(x)\,dx=3

Find: \displaystyle\int_3^{-2}f(x)\,dx if \displaystyle\int_{-2}^1f(x)\,dx=1 and \displaystyle\int_1^3f(x)\,dx=-5.

Solution

Method: Property (iii), then Property (ii)

The two given integrals join end to end: from -2 to 1 and from 1 to 3. Add them to get the integral from -2 to 3 (Property (iii)).
\int_{-2}^3f(x)\,dx=\int_{-2}^1f(x)\,dx+\int_1^3f(x)\,dx=1+(-5)=-4
The required integral has its limits the other way round (from 3 to -2). Interchanging the limits changes the sign of the integral (Property (ii)).
\int_3^{-2}f(x)\,dx=-\int_{-2}^3f(x)\,dx=-(-4)=4

\int_3^{-2}f(x)\,dx=4

Use appropriate formula from geometry to evaluate integrals.

Solution

Method. Draw the integrand, split the region into simple shapes (triangles, rectangles, parts of a circle) and add their signed areas.

(i) \displaystyle\int_{-1}^4(3-x)\,dx

The integrand is the line y=3-x with slope -1. It meets the x-axis at x=3. At the limits: x=-1\Rightarrow y=4 and x=4\Rightarrow y=-1.
From x=-1 to x=3 the line is above the axis. The region A_1 is a triangle with base 3-(-1)=4 and height 4.
A_1=\frac12(4)(4)=8
From x=3 to x=4 the line is below the axis. The region A_2 is a triangle with base 1 and height 1.
A_2=\frac12(1)(1)=\frac12
Add the signed areas. A_2 is below the axis, so subtract it.
\int_{-1}^4(3-x)\,dx=A_1-A_2=8-\frac12=\frac{15}{2}=7.5

A_1 is above the x-axis and A_2 is below it.

\int_{-1}^{4}(3-x)\,dx=\dfrac{15}{2}

(ii) \displaystyle\int_0^1\left[2+\sqrt{1-x^2}\right]dx

Split the integral of a sum into a sum of integrals (Theorem 3.3 (ii)).
\int_0^1\left[2+\sqrt{1-x^2}\right]dx=\int_0^1 2\,dx+\int_0^1\sqrt{1-x^2}\,dx
The first integral is a rectangle of width 1 and height 2.
\int_0^1 2\,dx=1\times2=2
For the second, y=\sqrt{1-x^2} means x^2+y^2=1 with y\ge0: the upper half of the circle of radius 1 centred at the origin. From x=0 to x=1 the region is a quarter circle (as in Example 20 (iii)).
\int_0^1\sqrt{1-x^2}\,dx=\frac14\pi(1)^2=\frac{\pi}{4}
Add the two parts.
\int_0^1\left[2+\sqrt{1-x^2}\right]dx=2+\frac{\pi}{4}\approx2.785

A 1\times2 rectangle with a quarter circle of radius 1 on top.

\int_0^1\left[2+\sqrt{1-x^2}\right]dx=2+\dfrac{\pi}{4}

(iii) \displaystyle\int_2^2\sqrt{x^3-4}\,dx

The lower and upper limits are equal. The region from x=2 to x=2 has zero width, so its area is zero (Property (i)).
\int_2^2\sqrt{x^3-4}\,dx=0
No formula for \sqrt{x^3-4} is needed. The integrand is defined at x=2, since 2^3-4=4\ge0.

\int_2^2\sqrt{x^3-4}\,dx=0

Exercise 3.7p. 112

Evaluate the definite integrals. Every question uses the Fundamental Theorem of Calculus: find any anti-derivative F(x), then work out F(b)-F(a) (upper limit first, then lower limit). No constant c is written, because it cancels. The book writes \left|F(x)\right|_{a}^{b}; these notes write \left[F(x)\right]_{a}^{b}, which means the same thing and cannot be confused with an absolute value.

Evaluate \displaystyle\int_{-1}^{2}(2x+3)\,dx.

Solution

Integrate each term with the power rule \int x^{n}dx=\dfrac{x^{n+1}}{n+1}. In a definite integral we do not write c, because it cancels in the subtraction.
\int_{-1}^{2}(2x+3)\,dx=\left[\dfrac{2x^{2}}{2}+3x\right]_{-1}^{2}=\left[x^{2}+3x\right]_{-1}^{2}
Apply the Fundamental Theorem: the value at the upper limit x=2 minus the value at the lower limit x=-1.
=\left(2^{2}+3(2)\right)-\left((-1)^{2}+3(-1)\right)=(4+6)-(1-3)
Simplify. Take care with the double negative.
=10-(-2)=12

\int_{-1}^{2}(2x+3)\,dx=12

Evaluate \displaystyle\int_{-4}^{12}\sqrt{y+4}\,dy.

Solution

Write the root as a power. The variable of integration is y, so we integrate with respect to y.
\int_{-4}^{12}\sqrt{y+4}\,dy=\int_{-4}^{12}(y+4)^{\frac12}\,dy
Use \int(ax+b)^{n}dx=\dfrac{(ax+b)^{n+1}}{a(n+1)} with a=1 and n=\tfrac12, so n+1=\tfrac32.
=\left[\dfrac{(y+4)^{\frac32}}{\frac32}\right]_{-4}^{12}=\dfrac23\left[(y+4)^{\frac32}\right]_{-4}^{12}
Substitute the limits: 12+4=16 and -4+4=0.
=\dfrac23\left(16^{\frac32}-0^{\frac32}\right)
Since 16^{3/2}=\left(\sqrt{16}\right)^{3}=4^{3}=64:
=\dfrac23(64-0)=\dfrac{128}{3}

\int_{-4}^{12}\sqrt{y+4}\,dy=\dfrac{128}{3}\approx42.67

Evaluate \displaystyle\int_{0}^{\frac12}(2x+1)^{-\frac13}\,dx.

Solution

Use \int(ax+b)^{n}dx=\dfrac{(ax+b)^{n+1}}{a(n+1)} with a=2 and n=-\tfrac13, so n+1=\tfrac23 and the divisor is 2\times\tfrac23=\tfrac43.
\int_{0}^{\frac12}(2x+1)^{-\frac13}dx=\left[\dfrac{(2x+1)^{\frac23}}{\frac43}\right]_{0}^{\frac12}=\dfrac34\left[(2x+1)^{\frac23}\right]_{0}^{\frac12}
Substitute the limits. At x=\tfrac12, 2x+1=2; at x=0, 2x+1=1.
=\dfrac34\left(2^{\frac23}-1^{\frac23}\right)=\dfrac34\left(\sqrt[3]{4}-1\right)
Give a decimal value too, using \sqrt[3]{4}\approx1.5874.
\approx\dfrac34(0.5874)=0.4406

\dfrac34\left(\sqrt[3]{4}-1\right)\approx0.441

Evaluate \displaystyle\int_{0}^{3}(6x^{2}-4x+5)\,dx.

Solution

Integrate term by term with the power rule.
\int_{0}^{3}(6x^{2}-4x+5)\,dx=\left[\dfrac{6x^{3}}{3}-\dfrac{4x^{2}}{2}+5x\right]_{0}^{3}=\left[2x^{3}-2x^{2}+5x\right]_{0}^{3}
Substitute the limits. Every term is zero at x=0, so only the upper limit contributes.
=\left(2(27)-2(9)+5(3)\right)-0=54-18+15
Add.
=51

\int_{0}^{3}(6x^{2}-4x+5)\,dx=51

Evaluate \displaystyle\int_{-2}^{1}(12x^{5}-36)\,dx.

Solution

Integrate each term with the power rule.
\int_{-2}^{1}(12x^{5}-36)\,dx=\left[\dfrac{12x^{6}}{6}-36x\right]_{-2}^{1}=\left[2x^{6}-36x\right]_{-2}^{1}
Substitute the limits. Note that (-2)^{6}=64 is positive (even power).
=\left(2(1)-36(1)\right)-\left(2(64)-36(-2)\right)=(2-36)-(128+72)
Simplify.
=-34-200=-234

\int_{-2}^{1}(12x^{5}-36)\,dx=-234

Evaluate \displaystyle\int_{-\pi/3}^{\pi/4}\cos\theta\,d\theta.

Solution

An anti-derivative of \cos\theta is \sin\theta.
\int_{-\pi/3}^{\pi/4}\cos\theta\,d\theta=\left[\sin\theta\right]_{-\pi/3}^{\pi/4}=\sin\dfrac{\pi}{4}-\sin\left(-\dfrac{\pi}{3}\right)
Sine is an odd function, \sin(-\theta)=-\sin\theta, so the second term becomes positive.
=\dfrac{1}{\sqrt2}+\dfrac{\sqrt3}{2}=\dfrac{\sqrt2}{2}+\dfrac{\sqrt3}{2}
Combine over the common denominator.
=\dfrac{\sqrt2+\sqrt3}{2}\approx1.573

\int_{-\pi/3}^{\pi/4}\cos\theta\,d\theta=\dfrac{\sqrt2+\sqrt3}{2}\approx1.573

Evaluate \displaystyle\int_{0}^{\pi/4}\sec^{2}2\theta\,d\theta.

Solution

Before using the Fundamental Theorem, check its condition: the integrand must be continuous on the whole interval [a,b].

Find where \sec^{2}2\theta is not defined. Since \sec2\theta=\dfrac{1}{\cos2\theta}, it fails when \cos2\theta=0.
\cos2\theta=0\;\Longrightarrow\;2\theta=\dfrac{\pi}{2}\;\Longrightarrow\;\theta=\dfrac{\pi}{4}
So the integrand is not defined at the upper limit \theta=\tfrac{\pi}{4}. It is not continuous on \left[0,\tfrac{\pi}{4}\right], and the theorem cannot be applied directly.
An anti-derivative is \tfrac12\tan2\theta (its derivative is \tfrac12\cdot2\sec^{2}2\theta). As \theta approaches \tfrac{\pi}{4} from below, \tan2\theta grows without limit.
\dfrac{d}{d\theta}\left(\dfrac12\tan2\theta\right)=\sec^{2}2\theta,\qquad \lim_{\theta\to\pi/4^{-}}\dfrac12\tan2\theta=+\infty
Conclusion: the region under the curve is infinitely large, so the integral as printed has no finite value. Writing \tfrac12\left(\tan\tfrac{\pi}{2}-\tan0\right) is wrong, because \tan\tfrac{\pi}{2} does not exist.

y=\sec^{2}2\theta shoots up without limit as \theta\to\tfrac{\pi}{4}, so the region under it from 0 to \tfrac{\pi}{4} has no finite area.

The likely intended questions

If the integrand is \sec^{2}\theta: it is continuous on \left[0,\tfrac{\pi}{4}\right] and an anti-derivative is \tan\theta.
\int_{0}^{\pi/4}\sec^{2}\theta\,d\theta=\left[\tan\theta\right]_{0}^{\pi/4}=\tan\dfrac{\pi}{4}-\tan0=1-0=1
If the upper limit is \tfrac{\pi}{8}: now 2\theta only runs from 0 to \tfrac{\pi}{4}, so \sec^{2}2\theta is continuous.
\int_{0}^{\pi/8}\sec^{2}2\theta\,d\theta=\left[\dfrac12\tan2\theta\right]_{0}^{\pi/8}=\dfrac12\left(\tan\dfrac{\pi}{4}-\tan0\right)=\dfrac12

\text{As printed: no finite value;}\quad \int_{0}^{\pi/4}\sec^{2}\theta\,d\theta=1,\quad \int_{0}^{\pi/8}\sec^{2}2\theta\,d\theta=\dfrac12

Note. Book misprint: \sec^{2}2\theta is not defined at \theta=\tfrac{\pi}{4}, so the printed integral does not exist (it diverges). The two likely intended versions are solved above.

Evaluate \displaystyle\int_{2}^{4}\dfrac{x^{2}+8}{x^{2}}\,dx.

Solution

Divide each term of the numerator by x^{2} so that the power rule can be used.
\dfrac{x^{2}+8}{x^{2}}=1+\dfrac{8}{x^{2}}=1+8x^{-2}
Integrate. For the second term, \int x^{-2}dx=\dfrac{x^{-1}}{-1}=-\dfrac1x.
\int_{2}^{4}\left(1+8x^{-2}\right)dx=\left[x-\dfrac{8}{x}\right]_{2}^{4}
Substitute the limits.
=\left(4-\dfrac84\right)-\left(2-\dfrac82\right)=(4-2)-(2-4)=2+2=4

\int_{2}^{4}\dfrac{x^{2}+8}{x^{2}}\,dx=4

Evaluate \displaystyle\int_{-1/2}^{3/2}(x-\cos\pi x)\,dx.

Solution

Integrate each term. For \cos\pi x use \int\cos ax\,dx=\dfrac{\sin ax}{a} with a=\pi.
\int_{-1/2}^{3/2}(x-\cos\pi x)\,dx=\left[\dfrac{x^{2}}{2}-\dfrac{\sin\pi x}{\pi}\right]_{-1/2}^{3/2}
Value at the upper limit x=\tfrac32, where \sin\tfrac{3\pi}{2}=-1.
\dfrac{(3/2)^{2}}{2}-\dfrac{\sin(3\pi/2)}{\pi}=\dfrac98+\dfrac1\pi
Value at the lower limit x=-\tfrac12, where \sin\left(-\tfrac{\pi}{2}\right)=-1.
\dfrac{(-1/2)^{2}}{2}-\dfrac{\sin(-\pi/2)}{\pi}=\dfrac18+\dfrac1\pi
Subtract. The \tfrac1\pi terms cancel.
\left(\dfrac98+\dfrac1\pi\right)-\left(\dfrac18+\dfrac1\pi\right)=\dfrac88=1

\int_{-1/2}^{3/2}(x-\cos\pi x)\,dx=1

Note. The book prints the integrand without brackets, as x-\cos\pi x\,dx. It means (x-\cos\pi x)\,dx.

Evaluate \displaystyle\int_{1}^{4}\dfrac{\cos\sqrt{x}}{2\sqrt{x}}\,dx.

Solution

The factor \dfrac{1}{2\sqrt x} is the derivative of \sqrt x, the expression inside the cosine. So substitute u=\sqrt x.
u=\sqrt x\;\Longrightarrow\;du=\dfrac{1}{2\sqrt x}\,dx
Change the limits (Method 2 of Example 25): x=1 gives u=1 and x=4 gives u=2.
\int_{1}^{4}\dfrac{\cos\sqrt x}{2\sqrt x}\,dx=\int_{1}^{2}\cos u\,du
Integrate and apply the new limits.
=\left[\sin u\right]_{1}^{2}=\sin2-\sin1
Evaluate with the calculator in radian mode.
\approx0.9093-0.8415=0.0678

\int_{1}^{4}\dfrac{\cos\sqrt x}{2\sqrt x}\,dx=\sin2-\sin1\approx0.068

Evaluate \displaystyle\int_{\pi/6}^{\pi/3}\sin x\cos x\,dx.

Solution

The derivative of \sin x is \cos x, so with u=\sin x the integral becomes \int u\,du=\dfrac{u^{2}}{2}.
\int\sin x\cos x\,dx=\dfrac{\sin^{2}x}{2}
Apply the limits, using \sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2} and \sin\tfrac{\pi}{6}=\tfrac12.
\int_{\pi/6}^{\pi/3}\sin x\cos x\,dx=\left[\dfrac{\sin^{2}x}{2}\right]_{\pi/6}^{\pi/3}=\dfrac12\left(\dfrac34-\dfrac14\right)
Simplify.
=\dfrac12\cdot\dfrac12=\dfrac14

\int_{\pi/6}^{\pi/3}\sin x\cos x\,dx=\dfrac14

Evaluate \displaystyle\int_{\pi/6}^{\pi/2}\dfrac{1+\cos\theta}{(\theta+\sin\theta)^{2}}\,d\theta.

Solution

The numerator 1+\cos\theta is the derivative of \theta+\sin\theta, the expression in the bracket. So substitute u=\theta+\sin\theta.
u=\theta+\sin\theta\;\Longrightarrow\;du=(1+\cos\theta)\,d\theta
Change the limits.
\theta=\dfrac\pi6:\ u=\dfrac\pi6+\dfrac12=\dfrac{\pi+3}{6},\qquad \theta=\dfrac\pi2:\ u=\dfrac\pi2+1=\dfrac{\pi+2}{2}
The integral becomes \int u^{-2}du=-\dfrac1u.
\int_{\frac{\pi+3}{6}}^{\frac{\pi+2}{2}}\dfrac{du}{u^{2}}=\left[-\dfrac1u\right]_{\frac{\pi+3}{6}}^{\frac{\pi+2}{2}}=-\dfrac{2}{\pi+2}+\dfrac{6}{\pi+3}
Combine over a common denominator.
=\dfrac{6(\pi+2)-2(\pi+3)}{(\pi+2)(\pi+3)}=\dfrac{4\pi+6}{(\pi+2)(\pi+3)}\approx0.588

\dfrac{6}{\pi+3}-\dfrac{2}{\pi+2}=\dfrac{4\pi+6}{(\pi+2)(\pi+3)}\approx0.588

Evaluate \displaystyle\int_{-\pi/4}^{\pi/4}(\sec x+\tan x)^{2}\,dx.

Solution

Expand the square.
(\sec x+\tan x)^{2}=\sec^{2}x+2\sec x\tan x+\tan^{2}x
Replace \tan^{2}x by \sec^{2}x-1 (from 1+\tan^{2}x=\sec^{2}x), so that every term has a standard integral.
=2\sec^{2}x+2\sec x\tan x-1
Integrate, using \int\sec^{2}x\,dx=\tan x and \int\sec x\tan x\,dx=\sec x.
\int_{-\pi/4}^{\pi/4}(\sec x+\tan x)^{2}dx=\left[2\tan x+2\sec x-x\right]_{-\pi/4}^{\pi/4}
Substitute, using \tan\left(\pm\tfrac\pi4\right)=\pm1 and \sec\left(\pm\tfrac\pi4\right)=\sqrt2.
=\left(2+2\sqrt2-\dfrac\pi4\right)-\left(-2+2\sqrt2+\dfrac\pi4\right)
Simplify. The 2\sqrt2 terms cancel.
=4-\dfrac\pi2\approx2.429

\int_{-\pi/4}^{\pi/4}(\sec x+\tan x)^{2}\,dx=4-\dfrac{\pi}{2}\approx2.429

Evaluate \displaystyle\int_{-\pi/2}^{\pi/2}\cos^{2}x\,dx.

Solution

\cos^{2}x has no direct formula, so use the identity from Section 3.2.1.
\cos^{2}x=\dfrac{1+\cos2x}{2}
Integrate each term.
\int_{-\pi/2}^{\pi/2}\cos^{2}x\,dx=\dfrac12\int_{-\pi/2}^{\pi/2}(1+\cos2x)\,dx=\dfrac12\left[x+\dfrac{\sin2x}{2}\right]_{-\pi/2}^{\pi/2}
Substitute. Both \sin\pi and \sin(-\pi) are 0.
=\dfrac12\left[\left(\dfrac\pi2+0\right)-\left(-\dfrac\pi2+0\right)\right]=\dfrac12(\pi)=\dfrac\pi2

\int_{-\pi/2}^{\pi/2}\cos^{2}x\,dx=\dfrac{\pi}{2}

Evaluate \displaystyle\int_{1}^{3}\ln x\,dx.

Solution

Write \ln x as (\ln x)(1) and integrate by parts. Take \ln x as the first function (its derivative \tfrac1x is simple) and 1 as the second function.
\int\ln x\,dx=\ln x\cdot x-\int\dfrac1x\cdot x\,dx
The new integral is simply \int1\,dx.
=x\ln x-\int1\,dx=x\ln x-x
Apply the limits. Remember \ln1=0.
\int_{1}^{3}\ln x\,dx=\left[x\ln x-x\right]_{1}^{3}=(3\ln3-3)-(1\cdot0-1)
Simplify.
=3\ln3-2\approx1.296

\int_{1}^{3}\ln x\,dx=3\ln3-2\approx1.296

Evaluate \displaystyle\int_{2}^{4}\left(e^{\frac x2}-e^{\frac x4}\right)dx.

Solution

Use \int e^{ax}dx=\dfrac{e^{ax}}{a}. With a=\tfrac12 this gives 2e^{x/2}, and with a=\tfrac14 it gives 4e^{x/4}.
\int_{2}^{4}\left(e^{\frac x2}-e^{\frac x4}\right)dx=\left[2e^{\frac x2}-4e^{\frac x4}\right]_{2}^{4}
Substitute the limits.
=\left(2e^{2}-4e^{1}\right)-\left(2e^{1}-4e^{\frac12}\right)
Collect like terms, writing e^{1/2}=\sqrt e.
=2e^{2}-6e+4\sqrt e\approx14.778-16.310+6.595=5.063

2e^{2}-6e+4\sqrt{e}\approx5.063

Evaluate \displaystyle\int_{0}^{\pi/4}\dfrac{1}{1-\sin x}\,dx.

Solution

Multiply the numerator and the denominator by 1+\sin x, as in Example 26. The denominator becomes 1-\sin^{2}x=\cos^{2}x.
\dfrac{1}{1-\sin x}\times\dfrac{1+\sin x}{1+\sin x}=\dfrac{1+\sin x}{1-\sin^{2}x}=\dfrac{1+\sin x}{\cos^{2}x}
Split the fraction into two standard integrands.
\dfrac{1}{\cos^{2}x}+\dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x}=\sec^{2}x+\sec x\tan x
Integrate.
\int_{0}^{\pi/4}\dfrac{dx}{1-\sin x}=\left[\tan x+\sec x\right]_{0}^{\pi/4}
Substitute: \tan\tfrac\pi4=1, \sec\tfrac\pi4=\sqrt2, \tan0=0, \sec0=1.
=(1+\sqrt2)-(0+1)=\sqrt2\approx1.414

\int_{0}^{\pi/4}\dfrac{dx}{1-\sin x}=\sqrt2

Evaluate \displaystyle\int_{0}^{\pi/4}\tan^{-1}y\,dy.

Solution

There is no direct formula for \tan^{-1}y. Write it as (\tan^{-1}y)(1) and integrate by parts, taking the inverse trigonometric function as the first function.
\int\tan^{-1}y\,dy=\tan^{-1}y\cdot y-\int\dfrac{1}{1+y^{2}}\cdot y\,dy
In the new integral the numerator y is half the derivative of 1+y^{2}.
\int\dfrac{y}{1+y^{2}}\,dy=\dfrac12\ln(1+y^{2})
So an anti-derivative is:
\int\tan^{-1}y\,dy=y\tan^{-1}y-\dfrac12\ln(1+y^{2})
Apply the limits. The upper limit is the number \tfrac\pi4\approx0.7854 (a value of y, not an angle). Both terms are 0 at y=0.
\int_{0}^{\pi/4}\tan^{-1}y\,dy=\dfrac\pi4\tan^{-1}\dfrac\pi4-\dfrac12\ln\left(1+\dfrac{\pi^{2}}{16}\right)
Evaluate with the calculator in radian mode, where \tan^{-1}(0.7854)=0.6658.
\approx0.7854(0.6658)-\dfrac12\ln(1.6169)=0.5229-0.2402=0.2827

\dfrac{\pi}{4}\tan^{-1}\dfrac{\pi}{4}-\dfrac12\ln\left(1+\dfrac{\pi^{2}}{16}\right)\approx0.283

Evaluate \displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{(2+\cos x)(5+\cos x)}\,dx.

Solution

The numerator \sin x is minus the derivative of \cos x. So substitute u=\cos x.
u=\cos x\;\Longrightarrow\;du=-\sin x\,dx
Change the limits: x=0 gives u=1 and x=\tfrac\pi2 gives u=0. The minus sign is used to swap the limits back into increasing order.
\int_{0}^{\pi/2}\dfrac{\sin x\,dx}{(2+\cos x)(5+\cos x)}=\int_{1}^{0}\dfrac{-du}{(2+u)(5+u)}=\int_{0}^{1}\dfrac{du}{(2+u)(5+u)}
Resolve into partial fractions. From 1=A(5+u)+B(2+u): put u=-2 to get A=\tfrac13, and u=-5 to get B=-\tfrac13.
\dfrac{1}{(2+u)(5+u)}=\dfrac13\left(\dfrac{1}{2+u}-\dfrac{1}{5+u}\right)
Integrate. Both 2+u and 5+u are positive on [0,1], so no absolute value signs are needed.
\dfrac13\left[\ln(2+u)-\ln(5+u)\right]_{0}^{1}=\dfrac13\left[\ln\dfrac{2+u}{5+u}\right]_{0}^{1}
Substitute and use \ln a-\ln b=\ln\tfrac ab.
=\dfrac13\left(\ln\dfrac36-\ln\dfrac25\right)=\dfrac13\ln\left(\dfrac12\times\dfrac52\right)=\dfrac13\ln\dfrac54\approx0.0744

\dfrac13\ln\dfrac54\approx0.074

Evaluate \displaystyle\int_{2}^{5}\dfrac{1}{x(x+1)}\,dx.

Solution

Resolve into partial fractions. From 1=A(x+1)+Bx: put x=0 to get A=1, and x=-1 to get B=-1.
\dfrac{1}{x(x+1)}=\dfrac1x-\dfrac{1}{x+1}
Integrate. On [2,5] both x and x+1 are positive, so no absolute value signs are needed.
\int_{2}^{5}\left(\dfrac1x-\dfrac{1}{x+1}\right)dx=\left[\ln x-\ln(x+1)\right]_{2}^{5}=\left[\ln\dfrac{x}{x+1}\right]_{2}^{5}
Substitute the limits and combine the logarithms.
=\ln\dfrac56-\ln\dfrac23=\ln\left(\dfrac56\times\dfrac32\right)=\ln\dfrac54\approx0.223

\int_{2}^{5}\dfrac{dx}{x(x+1)}=\ln\dfrac54\approx0.223

Exercise 3.8p. 117–118

Area: sketch the region, find where the curves meet, and decide which curve is on top. The area under y=f(x)\ge0 is \int_{a}^{b}f(x)\,dx; between two curves it is \int_{a}^{b}[f(x)-g(x)]\,dx with f on top; a piece below the x-axis is counted as positive. Volume of revolution (disc method): V=\pi\int_{a}^{b}y^{2}\,dx about the x-axis, V=\pi\int_{c}^{d}x^{2}\,dy about the y-axis. Questions 12 to 17 apply integrals to motion, work, surpluses and revenue.

Find the area of region bounded by the curve y=x^{2}, the x-axis, lines x=1 and x=3.

Solution

On [1,3] the curve y=x^{2} lies above the x-axis, so the area is the definite integral of y from x=1 to x=3.
A=\int_{1}^{3}y\,dx=\int_{1}^{3}x^{2}\,dx
Integrate with the power rule and apply the limits.
=\left[\dfrac{x^{3}}{3}\right]_{1}^{3}=\dfrac{27}{3}-\dfrac13=\dfrac{26}{3}

The shaded region under y=x^{2} from x=1 to x=3 has area \tfrac{26}{3}.

A=\dfrac{26}{3}\approx8.67\text{ sq. units}

Find the area under the curve y=\sqrt{6x+4} (above x-axis) from x=0 to x=2.

Solution

A square root is never negative, so the curve is above the x-axis on [0,2] and the area is:
A=\int_{0}^{2}\sqrt{6x+4}\,dx=\int_{0}^{2}(6x+4)^{\frac12}\,dx
Use \int(ax+b)^{n}dx=\dfrac{(ax+b)^{n+1}}{a(n+1)} with a=6 and n=\tfrac12. The divisor is 6\times\tfrac32=9.
=\left[\dfrac{(6x+4)^{\frac32}}{9}\right]_{0}^{2}
Substitute: 6(2)+4=16 and 6(0)+4=4. Then 16^{3/2}=64 and 4^{3/2}=8.
=\dfrac19\left(16^{\frac32}-4^{\frac32}\right)=\dfrac19(64-8)=\dfrac{56}{9}

The region under y=\sqrt{6x+4} between x=0 and x=2.

A=\dfrac{56}{9}\approx6.22\text{ sq. units}

Find the area of region bounded by the curve y^{2}=4x and line x=3.

Solution

y^{2}=4x is a parabola that opens to the right, with vertex at the origin. Solving for y gives an upper half and a lower half.
y=2\sqrt x\ \ (\text{upper half}),\qquad y=-2\sqrt x\ \ (\text{lower half})
The line x=3 cuts the parabola where y^{2}=12, that is at (3,\pm2\sqrt3). So the region runs from x=0 to x=3.
y=\pm\sqrt{12}=\pm2\sqrt3
The parabola is symmetric about the x-axis, so the area is twice the area of the upper half.
A=2\int_{0}^{3}2\sqrt x\,dx=4\int_{0}^{3}x^{\frac12}\,dx
Integrate and apply the limits, using 3^{3/2}=3\sqrt3.
=4\left[\dfrac23x^{\frac32}\right]_{0}^{3}=\dfrac83\left(3\sqrt3\right)=8\sqrt3\approx13.86

The region inside the parabola y^{2}=4x and to the left of x=3. It is symmetric about the x-axis.

A=8\sqrt3\approx13.86\text{ sq. units}

In the figure, a sketch of the function y=\tfrac12(0.2x^{2}+x) is shown. Find: (i) the area of region A. (ii) the area of region B. (iii) area of the region from x=1 to x=4. (iv) area of the region from x=-1 to x=-4.

Accurate graph of y=\tfrac12(0.2x^{2}+x). Region A is from x=1 to x=3, region B from x=3 to x=4; region (iv) lies below the x-axis.

Solution

Simplify the function first and find one anti-derivative F(x) that every part can use.

y=\dfrac12(0.2x^{2}+x)=\dfrac{x^{2}}{10}+\dfrac{x}{2},\qquad F(x)=\int y\,dx=\dfrac{x^{3}}{30}+\dfrac{x^{2}}{4}

(i) the area of region A.

Region A lies above the x-axis between the lines x=1 and x=3.
A=\int_{1}^{3}\left(\dfrac{x^{2}}{10}+\dfrac x2\right)dx=\left[\dfrac{x^{3}}{30}+\dfrac{x^{2}}{4}\right]_{1}^{3}
Substitute the limits.
=\left(\dfrac{27}{30}+\dfrac94\right)-\left(\dfrac{1}{30}+\dfrac14\right)=\dfrac{26}{30}+2=\dfrac{43}{15}\approx2.87

A=\dfrac{43}{15}\approx2.87\text{ sq. units}

(ii) the area of region B.

Region B lies above the x-axis from x=3 to x=4. Use the same F(x).
B=\left[\dfrac{x^{3}}{30}+\dfrac{x^{2}}{4}\right]_{3}^{4}=\left(\dfrac{64}{30}+4\right)-\left(\dfrac{27}{30}+\dfrac94\right)
Simplify over the common denominator 60.
=\dfrac{37}{30}+\dfrac74=\dfrac{74+105}{60}=\dfrac{179}{60}\approx2.98

B=\dfrac{179}{60}\approx2.98\text{ sq. units}

(iii) area of the region from x=1 to x=4.

From x=1 to x=4 the curve stays above the x-axis, so this area is region A plus region B.
\int_{1}^{4}y\,dx=\left[\dfrac{x^{3}}{30}+\dfrac{x^{2}}{4}\right]_{1}^{4}=\left(\dfrac{64}{30}+4\right)-\left(\dfrac{1}{30}+\dfrac14\right)
Simplify, and check against parts (i) and (ii).
=\dfrac{21}{10}+\dfrac{15}{4}=\dfrac{117}{20}=5.85,\qquad \dfrac{43}{15}+\dfrac{179}{60}=\dfrac{351}{60}=\dfrac{117}{20}\ \checkmark

\dfrac{117}{20}=5.85\text{ sq. units}

(iv) area of the region from x=-1 to x=-4.

Find where the curve meets the x-axis.
\dfrac12(0.2x^{2}+x)=0\;\Longrightarrow\;x(0.2x+1)=0\;\Longrightarrow\;x=0\ \text{or}\ x=-5
Between x=-5 and x=0 the curve is below the x-axis (for example y=-0.6 at x=-2). So the integral from -4 to -1 will come out negative.
\int_{-4}^{-1}y\,dx=\left[\dfrac{x^{3}}{30}+\dfrac{x^{2}}{4}\right]_{-4}^{-1}=\left(-\dfrac{1}{30}+\dfrac14\right)-\left(-\dfrac{64}{30}+4\right)
Simplify.
=\dfrac{63}{30}-\dfrac{15}{4}=\dfrac{42-75}{20}=-\dfrac{33}{20}
Area cannot be negative, so take the absolute value.
\text{Area}=\left|-\dfrac{33}{20}\right|=\dfrac{33}{20}=1.65

\dfrac{33}{20}=1.65\text{ sq. units}

Note. The book's sketch is not to scale: y=\tfrac12(0.2x^{2}+x) actually meets the x-axis at x=-5 and x=0, and its height at x=4 is 3.6. The graph above is drawn accurately. As in the book's figure, region A is taken from x=1 to x=3 and region B from x=3 to x=4.

Find the area bounded by the graph: (i) y=1+\cos x; [0,3\pi] (ii) y=-1+\sin x; \left[-\tfrac{3\pi}{2},\tfrac{3\pi}{2}\right]

Solution

(i) y=1+\cos x; [0,3\pi]

Check the sign first. Since \cos x\ge-1, we have y=1+\cos x\ge0 for every x. The graph never goes below the x-axis (it only touches it at x=\pi and x=3\pi), so one integral gives the area.
A=\int_{0}^{3\pi}(1+\cos x)\,dx
Integrate: \int\cos x\,dx=\sin x.
=\left[x+\sin x\right]_{0}^{3\pi}=(3\pi+\sin3\pi)-(0+\sin0)
Since \sin3\pi=0 and \sin0=0:
=3\pi\approx9.42

y=1+\cos x stays on or above the x-axis, touching it at x=\pi and x=3\pi.

A=3\pi\approx9.42\text{ sq. units}

(ii) y=-1+\sin x; \left[-\tfrac{3\pi}{2},\tfrac{3\pi}{2}\right]

Check the sign. Since \sin x\le1, we have y=-1+\sin x\le0 for every x. The whole region lies below the x-axis, so the area is the absolute value of the integral.
A=\left|\int_{-3\pi/2}^{3\pi/2}(-1+\sin x)\,dx\right|=\int_{-3\pi/2}^{3\pi/2}(1-\sin x)\,dx
Integrate: \int(1-\sin x)\,dx=x+\cos x.
=\left[x+\cos x\right]_{-3\pi/2}^{3\pi/2}=\left(\dfrac{3\pi}{2}+\cos\dfrac{3\pi}{2}\right)-\left(-\dfrac{3\pi}{2}+\cos\left(-\dfrac{3\pi}{2}\right)\right)
Since \cos\left(\pm\tfrac{3\pi}{2}\right)=0:
=\dfrac{3\pi}{2}+\dfrac{3\pi}{2}=3\pi\approx9.42

y=-1+\sin x stays on or below the x-axis, touching it at x=-\tfrac{3\pi}{2} and x=\tfrac{\pi}{2}.

A=3\pi\approx9.42\text{ sq. units}

Note. The book prints 'bunded'; it means 'bounded'. In both parts the region is the one between the graph and the x-axis over the given interval.

Find the area of the region bounded by the graphs of y=x, y=-2x and x=3.

Solution

Find where the two lines meet: x=-2x gives x=0, so they meet at the origin. The region is the triangle between the lines from x=0 to x=3.
x=-2x\;\Longrightarrow\;3x=0\;\Longrightarrow\;x=0
For 0<x\le3 the line y=x is above y=-2x (at x=1: 1>-2). The area between curves is the integral of (upper curve minus lower curve).
A=\int_{0}^{3}\left[x-(-2x)\right]dx=\int_{0}^{3}3x\,dx
Integrate.
=\left[\dfrac{3x^{2}}{2}\right]_{0}^{3}=\dfrac{27}{2}=13.5
Check with geometry: the triangle has vertices (0,0), (3,3) and (3,-6). Its base on x=3 is 3-(-6)=9 and its height is 3.
\tfrac12\times9\times3=\dfrac{27}{2}\ \checkmark

The triangle bounded by y=x (top), y=-2x (bottom) and x=3.

A=\dfrac{27}{2}=13.5\text{ sq. units}

Find the area of the region bounded above by y=x+6, bounded below by y=x^{2} and bounded on the sides by the lines x=0 and x=2.

Solution

On [0,2] the line is above the parabola (at x=2: 8>4), so the height of a thin vertical strip is (x+6)-x^{2}.
A=\int_{0}^{2}\left[(x+6)-x^{2}\right]dx
Integrate term by term.
=\left[\dfrac{x^{2}}{2}+6x-\dfrac{x^{3}}{3}\right]_{0}^{2}=\left(2+12-\dfrac83\right)-0
Simplify.
=14-\dfrac83=\dfrac{34}{3}\approx11.33

The region between y=x+6 (top) and y=x^{2} (bottom) from the y-axis (x=0) to x=2.

A=\dfrac{34}{3}\approx11.33\text{ sq. units}

Find the area bounded by the curve y=x^{3}+1, the x-axis and the line x=1.

Solution

Find where the curve meets the x-axis.
x^{3}+1=0\;\Longrightarrow\;x^{3}=-1\;\Longrightarrow\;x=-1
The region enclosed by the curve, the x-axis and the line x=1 runs from x=-1 to x=1. There x^{3}\ge-1, so the curve is on or above the x-axis.
A=\int_{-1}^{1}(x^{3}+1)\,dx
Integrate and apply the limits.
=\left[\dfrac{x^{4}}{4}+x\right]_{-1}^{1}=\left(\dfrac14+1\right)-\left(\dfrac14-1\right)=2

The region under y=x^{3}+1 from its x-intercept x=-1 to the line x=1.

A=2\text{ sq. units}

Find the area of the region enclosed by x=y^{2} and y=x-2 integrating with respect to y.

Solution

Write both boundaries as x in terms of y: the parabola x=y^{2} and the line x=y+2.
x=y^{2},\qquad x=y+2
Find where they meet by equating the x-values.
y^{2}=y+2\;\Longrightarrow\;y^{2}-y-2=0\;\Longrightarrow\;(y-2)(y+1)=0\;\Longrightarrow\;y=-1,\ 2
The meeting points are (1,-1) and (4,2). For -1<y<2 the line is to the right of the parabola (at y=0: 2>0). With horizontal strips, the length of a strip is (right curve minus left curve).
A=\int_{-1}^{2}\left[(y+2)-y^{2}\right]dy
Integrate with respect to y.
=\left[\dfrac{y^{2}}{2}+2y-\dfrac{y^{3}}{3}\right]_{-1}^{2}=\left(2+4-\dfrac83\right)-\left(\dfrac12-2+\dfrac13\right)
Simplify.
=\dfrac{10}{3}-\left(-\dfrac76\right)=\dfrac{20}{6}+\dfrac76=\dfrac{27}{6}=\dfrac92

The region between the parabola x=y^{2} (left) and the line x=y+2 (right), for y from -1 to 2.

A=\dfrac92=4.5\text{ sq. units}

Find the volume of the solid that is obtained when the region under the curve y=\sqrt{x} over the interval [1,4] is revolved about the x-axis.

Solution

Each thin vertical strip of the region sweeps out a disc of radius y=\sqrt x and thickness dx, with volume \pi y^{2}\,dx. Adding the discs gives the disc method.
V=\pi\int_{a}^{b}y^{2}\,dx=\pi\int_{1}^{4}\left(\sqrt x\right)^{2}dx
Square the radius: \left(\sqrt x\right)^{2}=x.
=\pi\int_{1}^{4}x\,dx=\pi\left[\dfrac{x^{2}}{2}\right]_{1}^{4}
Substitute the limits.
=\pi\left(\dfrac{16}{2}-\dfrac12\right)=\dfrac{15\pi}{2}\approx23.56

Sketch: the shaded region R turns about the x-axis. The dashed curve (the mirror image y=-\sqrt x) and the ellipses show the solid it sweeps out.

V=\dfrac{15\pi}{2}\approx23.56\text{ cu. units}

Find the volume of the solid that results when the shaded region is revolved about the indicated axis. (i) y=\sqrt{3-x} about x-axis (ii) y=3-2x about y-axis

Solution

(i) y=\sqrt{3-x} about the x-axis (shaded region from x=-1 to x=3).

Read the region from the book's figure: it lies under y=\sqrt{3-x} from the line x=-1 to x=3, where the curve meets the x-axis.
\sqrt{3-x}=0\;\Longrightarrow\;x=3
Disc method about the x-axis, with radius y=\sqrt{3-x}.
V=\pi\int_{-1}^{3}\left(\sqrt{3-x}\right)^{2}dx=\pi\int_{-1}^{3}(3-x)\,dx
Integrate.
=\pi\left[3x-\dfrac{x^{2}}{2}\right]_{-1}^{3}=\pi\left[\left(9-\dfrac92\right)-\left(-3-\dfrac12\right)\right]
Simplify.
=\pi\left(\dfrac92+\dfrac72\right)=8\pi\approx25.13

Sketch: region R under y=\sqrt{3-x} from x=-1 to x=3 turns about the x-axis. The solid has a circular face of radius 2 at x=-1 and narrows to a point at x=3.

V=8\pi\approx25.13\text{ cu. units}

(ii) y=3-2x about the y-axis (shaded triangle between the line and the axes).

Read the region from the book's figure: the triangle between the line y=3-2x, the x-axis and the y-axis. The line meets the axes at (1.5,0) and (0,3), so y runs from 0 to 3.
y=0\Rightarrow x=1.5,\qquad x=0\Rightarrow y=3
The region turns about the y-axis, so use horizontal discs of radius x and thickness dy. Write x in terms of y.
y=3-2x\;\Longrightarrow\;x=g(y)=\dfrac{3-y}{2}
Disc method about the y-axis, V=\pi\int_{c}^{d}[g(y)]^{2}\,dy, as in Example 31.
V=\pi\int_{0}^{3}\dfrac{(3-y)^{2}}{4}\,dy
Integrate, using \int(3-y)^{2}dy=-\dfrac{(3-y)^{3}}{3}.
=\dfrac\pi4\left[-\dfrac{(3-y)^{3}}{3}\right]_{0}^{3}=\dfrac\pi4\left(0+\dfrac{27}{3}\right)=\dfrac{9\pi}{4}\approx7.07
Check: the solid is a cone with base radius r=1.5 and height h=3.
\tfrac13\pi r^{2}h=\tfrac13\pi(1.5)^{2}(3)=\dfrac{9\pi}{4}\ \checkmark

Sketch: the triangle R turns about the y-axis and sweeps out a cone of base radius 1.5 and height 3.

V=\dfrac{9\pi}{4}\approx7.07\text{ cu. units}

Note. The book prints 'sloid'; it means 'solid'. The regions are read from the book's figures.

An object moves in a straight line according to the position function given below. If S is measured in centimetres, find the distance travelled by the object in the indicated time interval: (i) S(t)=t^{2}-2t; [0,5] (ii) S(t)=t^{3}-3t^{2}-9t; [0,4] (iii) S(t)=6\sin\pi t; [1,3]

Solution

The distance travelled is not always S(b)-S(a); that difference is the displacement. If the object turns back during the interval, split the interval where the velocity v(t)=\dfrac{dS}{dt} is zero, and add the sizes of the pieces. On each piece, \int v\,dt is just the change in S.

\text{distance}=\int_{a}^{b}\left|v(t)\right|dt

(i) S(t)=t^{2}-2t; [0,5]

Velocity: differentiate the position.
v(t)=\dfrac{dS}{dt}=2t-2
v=0 at t=1. For 0\le t<1, v<0 (moving backwards); for 1<t\le5, v>0 (moving forwards). Split the interval at t=1.
\text{distance}=\left|\int_{0}^{1}(2t-2)\,dt\right|+\int_{1}^{5}(2t-2)\,dt
Integrate each piece, using \int(2t-2)\,dt=t^{2}-2t.
\left[t^{2}-2t\right]_{0}^{1}=-1,\qquad \left[t^{2}-2t\right]_{1}^{5}=15-(-1)=16
Add the sizes.
\text{distance}=|-1|+16=17\,\mathrm{cm}

Distance is the total shaded area between the velocity graph and the t-axis: A_1+A_2=1+16=17\,\mathrm{cm}.

17\,\mathrm{cm}

(ii) S(t)=t^{3}-3t^{2}-9t; [0,4]

Velocity: differentiate and factorise.
v(t)=3t^{2}-6t-9=3(t^{2}-2t-3)=3(t-3)(t+1)
In [0,4], v=0 only at t=3. For 0\le t<3, v<0; for 3<t\le4, v>0. Split at t=3 and use S(t) itself as the anti-derivative of v.
\int_{0}^{3}v\,dt=S(3)-S(0)=(27-27-27)-0=-27
Second piece.
\int_{3}^{4}v\,dt=S(4)-S(3)=(64-48-36)-(-27)=-20+27=7
Add the sizes.
\text{distance}=|-27|+7=34\,\mathrm{cm}

A_1=27 (below the axis: moving backwards) and A_2=7 (moving forwards). Distance =27+7=34\,\mathrm{cm}.

34\,\mathrm{cm}

(iii) S(t)=6\sin\pi t; [1,3]

Velocity: differentiate, using the chain rule.
v(t)=6\pi\cos\pi t
v=0 when \cos\pi t=0, that is \pi t=\tfrac{3\pi}{2} or \tfrac{5\pi}{2}. So t=\tfrac32 and t=\tfrac52 in [1,3]. The sign of v changes at each, so split into three pieces.
[1,3]=\left[1,\tfrac32\right]\cup\left[\tfrac32,\tfrac52\right]\cup\left[\tfrac52,3\right]
On each piece \int v\,dt is the change in S. Find S at the break points.
\begin{aligned}S(1)&=6\sin\pi=0\\ S\left(\tfrac32\right)&=6\sin\tfrac{3\pi}{2}=-6\\ S\left(\tfrac52\right)&=6\sin\tfrac{5\pi}{2}=6\\ S(3)&=6\sin3\pi=0\end{aligned}
Add the sizes of the three changes.
\text{distance}=|-6-0|+|6-(-6)|+|0-6|=6+12+6=24\,\mathrm{cm}

The velocity changes sign at t=1.5 and t=2.5. Distance =A_1+A_2+A_3=6+12+6=24\,\mathrm{cm}.

24\,\mathrm{cm}

Note. The book writes 'If f is measured in centimetres'; the position function is S.

It takes a force of 50\,\mathrm{N} to stretch a spring of 0.5\,\mathrm{m}. Find the work done in stretching the spring 0.6\,\mathrm{m} beyond its natural length.

Solution

Given

F = 50\,\mathrm{N} force needed for the first stretch
x = 0.5\,\mathrm{m} stretch produced by that force
x_{2} = 0.6\,\mathrm{m} required stretch beyond the natural length

To find

work done W

Hooke's law: F=kx, where x is the stretch beyond the natural length and k is the spring constant. Use the given pair to find k.
50=k(0.5)\;\Longrightarrow\;k=100\,\mathrm{N\,m^{-1}},\qquad F(x)=100x
The work done by a variable force from x=a to x=b is W=\int_{a}^{b}F(x)\,dx. The spring is stretched from its natural length (x=0) to x=0.6\,\mathrm{m}.
W=\int_{0}^{0.6}100x\,dx=\left[50x^{2}\right]_{0}^{0.6}
Substitute the limits.
=50(0.6)^{2}=50(0.36)=18\,\mathrm{J}

Work is the area under the force-extension line: W=\tfrac12(0.6)(60)=18\,\mathrm{J}.

W=18\,\mathrm{J}

Note. 'A force of 50\,\mathrm{N} to stretch a spring of 0.5\,\mathrm{m}' is read, as in Example 34, as: a force of 50\,\mathrm{N} stretches the spring 0.5\,\mathrm{m} beyond its natural length.

A force F=\tfrac32x\,\mathrm{lb} is needed to stretch a 10 inch spring an additional x inch. Find the work done in stretching the spring 16 inch.

Solution

The natural length is 10 inches and x is the extra stretch. Stretching the spring to a length of 16 inches means a stretch of 16-10=6 inches, so x runs from 0 to 6.
x:\ 0\ \to\ 16-10=6\ \text{inches}
Work done by the variable force is the integral of F over the stretch.
W=\int_{0}^{6}\dfrac32x\,dx=\left[\dfrac34x^{2}\right]_{0}^{6}=\dfrac34(36)=27\,\text{in-lb}
Convert to foot-pounds if needed (12 inches =1 foot).
27\,\text{in-lb}=\dfrac{27}{12}=2.25\,\text{ft-lb}

The work is the triangle under F=\tfrac32x up to x=6: W=\tfrac12(6)(9)=27 in-lb.

W=27\,\text{in-lb}=2.25\,\text{ft-lb}

Find the consumer and producer surpluses, when: (i) S(x)=24, D(x)=100-2x (ii) S(x)=x^{2}-4, D(x)=-x+8 (iii) S(x)=2x^{2}+3x, D(x)=36-x^{2}

Solution

Method (Section 3.12.1): x is the price per unit, D(x) is the number of units demanded and S(x) the number supplied at that price. Find the equilibrium price p from D(p)=S(p), the price b at which demand is zero, D(b)=0, and the price c at which supply is zero, S(c)=0. Then:

\begin{aligned}C_s&=\int_{p}^{b}D(x)\,dx\ \ (\text{consumer surplus})\\ P_s&=\int_{c}^{p}S(x)\,dx\ \ (\text{producer surplus})\end{aligned}

(i) S(x)=24, D(x)=100-2x

Equilibrium price: set demand equal to supply.
100-2p=24\;\Longrightarrow\;2p=76\;\Longrightarrow\;p=38
Price at which demand is zero.
100-2b=0\;\Longrightarrow\;b=50
Consumer surplus: area under D(x) from p=38 to b=50.
C_s=\int_{38}^{50}(100-2x)\,dx=\left[100x-x^{2}\right]_{38}^{50}=(5000-2500)-(3800-1444)=144
Producer surplus: here the supply is 24 units at every price, so it never becomes zero. The lowest possible price is 0, so we take c=0.
P_s=\int_{0}^{38}24\,dx=\left[24x\right]_{0}^{38}=912

x is the price. C_s is the area under D(x) from p=38 to b=50; P_s is the area under S(x) from 0 to 38.

C_s=144,\qquad P_s=912

(ii) S(x)=x^{2}-4, D(x)=-x+8

Equilibrium price. A price cannot be negative, so reject p=-4.
p^{2}-4=-p+8\;\Longrightarrow\;p^{2}+p-12=0\;\Longrightarrow\;(p+4)(p-3)=0\;\Longrightarrow\;p=3
Zero demand: -b+8=0 gives b=8. Zero supply: c^{2}-4=0 gives c=2 (the positive root).
b=8,\qquad c=2
Consumer surplus.
C_s=\int_{3}^{8}(8-x)\,dx=\left[8x-\dfrac{x^{2}}{2}\right]_{3}^{8}=(64-32)-\left(24-\dfrac92\right)=\dfrac{25}{2}=12.5
Producer surplus.
P_s=\int_{2}^{3}(x^{2}-4)\,dx=\left[\dfrac{x^{3}}{3}-4x\right]_{2}^{3}=(9-12)-\left(\dfrac83-8\right)=\dfrac73\approx2.33

Equilibrium at price p=3. C_s lies under D(x) from 3 to 8; P_s lies under S(x) from 2 to 3.

C_s=\dfrac{25}{2}=12.5,\qquad P_s=\dfrac73\approx2.33

(iii) S(x)=2x^{2}+3x, D(x)=36-x^{2}

Equilibrium price. Reject the negative root p=-4.
2p^{2}+3p=36-p^{2}\;\Longrightarrow\;3p^{2}+3p-36=0\;\Longrightarrow\;p^{2}+p-12=0\;\Longrightarrow\;p=3
Zero demand: 36-b^{2}=0 gives b=6. Zero supply: c(2c+3)=0 gives c=0 (the other root, -\tfrac32, is a negative price).
b=6,\qquad c=0
Consumer surplus.
C_s=\int_{3}^{6}(36-x^{2})\,dx=\left[36x-\dfrac{x^{3}}{3}\right]_{3}^{6}=(216-72)-(108-9)=144-99=45
Producer surplus.
P_s=\int_{0}^{3}(2x^{2}+3x)\,dx=\left[\dfrac{2x^{3}}{3}+\dfrac{3x^{2}}{2}\right]_{0}^{3}=18+\dfrac{27}{2}=\dfrac{63}{2}=31.5

Equilibrium at price p=3. C_s lies under D(x) from 3 to 6; P_s lies under S(x) from 0 to 3.

C_s=45,\qquad P_s=\dfrac{63}{2}=31.5

Note. Part (i): the book's rule needs a price c with S(c)=0, but the constant supply S(x)=24 is never zero, so the lowest price c=0 is used. This is an assumption; the answers follow the book's method of Example 32.

Find the total revenue obtained in 4 years if the rate of increase in dollars per year is: f(t)=200(t-5)^{2}

Solution

The rate of change of revenue is \dfrac{dR}{dt}=f(t) dollars per year. So the total revenue in the first 4 years is the integral of the rate from t=0 to t=4.
R=\int_{0}^{4}200(t-5)^{2}\,dt
Use \int(t-5)^{2}dt=\dfrac{(t-5)^{3}}{3}.
=200\left[\dfrac{(t-5)^{3}}{3}\right]_{0}^{4}=\dfrac{200}{3}\left[(-1)^{3}-(-5)^{3}\right]
Simplify. Note (-5)^{3}=-125, so subtracting it adds 125.
=\dfrac{200}{3}(-1+125)=\dfrac{200\times124}{3}=\dfrac{24800}{3}\approx8266.67

Total revenue is the area under the rate curve from t=0 to t=4.

R=\dfrac{24800}{3}\approx\$\,8266.67

Find the total revenue obtained in 8 years if the rate of increase in dollars per year is: f(t)=600\sqrt{1+3t}

Solution

Total revenue in the first 8 years is the integral of the rate from t=0 to t=8.
R=\int_{0}^{8}600\sqrt{1+3t}\,dt=600\int_{0}^{8}(1+3t)^{\frac12}\,dt
Use \int(at+b)^{n}dt=\dfrac{(at+b)^{n+1}}{a(n+1)} with a=3 and n=\tfrac12: divide by 3\times\tfrac32=\tfrac92.
=600\left[\dfrac{2(1+3t)^{\frac32}}{9}\right]_{0}^{8}=\dfrac{400}{3}\left[(1+3t)^{\frac32}\right]_{0}^{8}
Substitute: 1+3(8)=25 and 25^{3/2}=5^{3}=125; 1+3(0)=1.
=\dfrac{400}{3}(125-1)=\dfrac{49600}{3}\approx16533.33

Total revenue is the area under the rate curve from t=0 to t=8.

R=\dfrac{49600}{3}\approx\$\,16533.33

Find the area bounded by the curve f(x)=x^{3}-2x^{2}+1 and the x-axis in the first quadrant bounded by the line x=1.5.

Solution

Find where the curve meets the x-axis. x=1 is a root because 1-2+1=0, so (x-1) is a factor.
x^{3}-2x^{2}+1=(x-1)(x^{2}-x-1)=0\;\Longrightarrow\;x=1,\ \ x=\dfrac{1\pm\sqrt5}{2}
From x=0 to x=1.5 the only root is x=1 (the others are about -0.618 and 1.618). The curve is above the x-axis on [0,1] and below it on [1,1.5] (for example f(1.5)=-0.125). So split the integral at x=1.
A=\int_{0}^{1}f(x)\,dx+\left|\int_{1}^{1.5}f(x)\,dx\right|
Anti-derivative.
F(x)=\dfrac{x^{4}}{4}-\dfrac{2x^{3}}{3}+x
First part.
\int_{0}^{1}f(x)\,dx=F(1)-F(0)=\dfrac14-\dfrac23+1=\dfrac{7}{12}
Second part: F(1.5)=\dfrac{5.0625}{4}-\dfrac{2(3.375)}{3}+1.5=0.515625=\dfrac{33}{64}.
\int_{1}^{1.5}f(x)\,dx=\dfrac{33}{64}-\dfrac{7}{12}=\dfrac{99-112}{192}=-\dfrac{13}{192}
Add the sizes of the two parts.
A=\dfrac{7}{12}+\dfrac{13}{192}=\dfrac{112+13}{192}=\dfrac{125}{192}\approx0.651

A_1=\tfrac{7}{12} lies above the x-axis; the thin piece A_2=\tfrac{13}{192} lies just below it.

A=\dfrac{125}{192}\approx0.651\text{ sq. units}

Note. The book prints 'bonded'; it means 'bounded'. The piece from x=1 to x=1.5 lies just below the x-axis. If only the part inside the first quadrant is wanted, the area is \tfrac{7}{12}\approx0.583 sq. units.

Review Exercisep. 119

Question 1 revises the basic rules and the Fundamental Theorem. Question 2 mixes every method of the unit: first decide the method (standard form, substitution, trigonometric identity, partial fractions or integration by parts), then integrate. Questions 4 and 5 use v=\int a\,dt and S=\int v\,dt with the starting values to fix the constants.

1 (i)

If f is integrable, then it is:

Adiscontinuous
Bunbounded
CcontinuousAnswer
Dlinear

Why

The book's Theorem 3.2 links the two ideas: a function that is continuous on [a,b] is integrable on [a,b], and every function integrated in this unit is continuous, so the expected answer is (c). An unbounded function is not integrable, so (b) is wrong, and being discontinuous or linear is not required.

Note. Strictly, the theorem works in one direction: continuous implies integrable. An integrable function must be bounded but need not be continuous (a step function is integrable). Option (c) is the answer the book expects.

1 (ii)

If f'(x)=3x^{2}+2x, then f(x) is:

A6x+2+c
Bx^{3}+x^{2}+cAnswer
C3x^{3}+2x^{2}+c
D1.5x^{3}+x^{2}+c

Why

Integrate f'(x) term by term: \int(3x^{2}+2x)\,dx=\dfrac{3x^{3}}{3}+\dfrac{2x^{2}}{2}+c=x^{3}+x^{2}+c. Check: the derivative of x^{3}+x^{2} is 3x^{2}+2x. Option (a) is the derivative of f'(x), not its integral.

1 (iii)

\displaystyle\int\dfrac{d}{dx}(x^{2})\,dx is equal to:

Ax^{2}+cAnswer
B2x+c
C\dfrac{x^{3}}{3}+c
D2x+c

Why

Integration undoes differentiation: \int\dfrac{d}{dx}(x^{2})\,dx=\int2x\,dx=x^{2}+c.

Note. Options (b) and (d) are printed the same (2x+c); this is a misprint in the book, and neither is correct.

1 (iv)

\displaystyle\int\sin2x\,dx is:

A\dfrac{\cos2x}{2}+c
B2\cos2x+c
C-\dfrac{\sin2x}{2}+c
D-\dfrac{\cos2x}{2}+cAnswer

Why

Use \int\sin ax\,dx=-\dfrac{\cos ax}{a}+c with a=2. Check: \dfrac{d}{dx}\left(-\dfrac{\cos2x}{2}\right)=\dfrac{2\sin2x}{2}=\sin2x. Option (a) has the wrong sign.

1 (v)

\displaystyle\int_{3}^{7}dx is:

A3
B4Answer
C5
D6

Why

\int_{3}^{7}dx=\left[x\right]_{3}^{7}=7-3=4. It is the area of a rectangle of height 1 and width 4.

1 (vi)

\displaystyle\int_{\pi/6}^{\pi}\cos x\,dx is:

A-\dfrac12Answer
B\dfrac12
C\dfrac32
D-\dfrac32

Why

\int_{\pi/6}^{\pi}\cos x\,dx=\left[\sin x\right]_{\pi/6}^{\pi}=\sin\pi-\sin\dfrac{\pi}{6}=0-\dfrac12=-\dfrac12. A definite integral can be negative: on this interval more of the cosine curve lies below the x-axis than above it.

1 (vii)

\dfrac{d}{dx}\displaystyle\int_{-2}^{x}t^{3}\,dt is equal to:

At^{4}
Bt^{3}
Cx^{3}Answer
Dx^{3}-16

Why

By the Fundamental Theorem, \dfrac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x). Directly: \int_{-2}^{x}t^{3}dt=\dfrac{x^{4}}{4}-4, whose derivative is x^{3}. The answer is in x, not t, so (a) and (b) are wrong.

1 (viii)

What is relation between \displaystyle\int_{1}^{2}x\,dx and \displaystyle\int_{1}^{2}t\,dt?

A\int_{1}^{2}x\,dx<\int_{1}^{2}t\,dt
B\int_{1}^{2}x\,dx>\int_{1}^{2}t\,dt
C\int_{1}^{2}x\,dx\neq\int_{1}^{2}t\,dt
D\int_{1}^{2}x\,dx=\int_{1}^{2}t\,dtAnswer

Why

The letter used for the variable of integration does not matter; it is a dummy variable. Both integrals equal \left[\tfrac{u^{2}}{2}\right]_{1}^{2}=2-\tfrac12=\tfrac32.

1 (ix)

Area under the graph of f(x)=4; [2,5] is:

A2
B4
C5
D12Answer

Why

\int_{2}^{5}4\,dx=4(5-2)=12. The region is a rectangle of height 4 and width 3.

The rectangle under y=4 from x=2 to x=5 has area 4\times3=12.

1 (x)

\displaystyle\int\sqrt{x}\,dx is:

Ax^{\frac32}+c
B\dfrac23x^{\frac32}+cAnswer
C\dfrac32x^{\frac32}+c
Dx^{\frac12}+c

Why

Power rule: \int x^{\frac12}dx=\dfrac{x^{\frac32}}{\frac32}+c=\dfrac23x^{\frac32}+c. Option (c) multiplies by \tfrac32 instead of dividing by it.

Evaluate: (i) \displaystyle\int\dfrac{4x+2}{x^{2}+x+1}dx (ii) \displaystyle\int x(x^{2}+1)^{4}dx (iii) \displaystyle\int\cos^{2}3x\,dx (iv) \displaystyle\int\dfrac{x^{2}-29x+5}{(x-4)^{2}(x^{2}+3)}dx (v) \displaystyle\int\sin^{-1}x\,dx (vi) \displaystyle\int2x\sin3x\,dx (vii) \displaystyle\int x^{2}e^{x}dx (viii) \displaystyle\int_{0}^{\pi/4}(\sin2x-5\cos4x)\,dx (ix) \displaystyle\int_{1}^{4}\dfrac{\cos\sqrt x}{2\sqrt x}dx

Solution

(i) \displaystyle\int\dfrac{4x+2}{x^{2}+x+1}dx

Compare the numerator with the derivative of the denominator. The numerator is exactly twice it.
\dfrac{d}{dx}(x^{2}+x+1)=2x+1,\qquad 4x+2=2(2x+1)
Use \int\dfrac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c.
\int\dfrac{4x+2}{x^{2}+x+1}dx=2\int\dfrac{2x+1}{x^{2}+x+1}dx=2\ln\left|x^{2}+x+1\right|+c
Since x^{2}+x+1=\left(x+\tfrac12\right)^{2}+\tfrac34>0 for every x, the absolute value signs can be dropped.
=2\ln(x^{2}+x+1)+c

2\ln(x^{2}+x+1)+c

(ii) \displaystyle\int x(x^{2}+1)^{4}dx

Substitute u=x^{2}+1, because its derivative 2x matches the factor x outside the bracket (up to the constant 2).
u=x^{2}+1\;\Longrightarrow\;du=2x\,dx\;\Longrightarrow\;x\,dx=\dfrac{du}{2}
Rewrite in terms of u and integrate.
\int x(x^{2}+1)^{4}dx=\dfrac12\int u^{4}\,du=\dfrac12\cdot\dfrac{u^{5}}{5}+c=\dfrac{u^{5}}{10}+c
Put back u=x^{2}+1.
=\dfrac{(x^{2}+1)^{5}}{10}+c

\dfrac{(x^{2}+1)^{5}}{10}+c

(iii) \displaystyle\int\cos^{2}3x\,dx

Use the identity \cos^{2}\theta=\dfrac{1+\cos2\theta}{2} with \theta=3x (Section 3.2.1).
\cos^{2}3x=\dfrac{1+\cos6x}{2}
Integrate term by term, with \int\cos6x\,dx=\dfrac{\sin6x}{6}.
\int\cos^{2}3x\,dx=\dfrac12\int(1+\cos6x)\,dx=\dfrac12\left(x+\dfrac{\sin6x}{6}\right)+c
Simplify.
=\dfrac{x}{2}+\dfrac{\sin6x}{12}+c

\dfrac{x}{2}+\dfrac{\sin6x}{12}+c

(iv) \displaystyle\int\dfrac{x^{2}-29x+5}{(x-4)^{2}(x^{2}+3)}dx

The denominator has a repeated linear factor (x-4)^{2} and a quadratic factor x^{2}+3 that cannot be factorised. Set up the partial fractions.
\dfrac{x^{2}-29x+5}{(x-4)^{2}(x^{2}+3)}=\dfrac{A}{x-4}+\dfrac{B}{(x-4)^{2}}+\dfrac{Cx+D}{x^{2}+3}
Multiply both sides by (x-4)^{2}(x^{2}+3).
x^{2}-29x+5=A(x-4)(x^{2}+3)+B(x^{2}+3)+(Cx+D)(x-4)^{2}
Put x=4: every term containing (x-4) vanishes.
16-116+5=19B\;\Longrightarrow\;-95=19B\;\Longrightarrow\;B=-5
Compare coefficients. x^{3}: 0=A+C. x^{2}: 1=-4A+B-8C+D. Constant: 5=-12A+3B+16D. Put B=-5.
A+C=0,\qquad -4A-8C+D=6,\qquad -12A+16D=20
Solve. C=-A turns the second equation into 4A+D=6; the third is -3A+4D=5. So D=6-4A and -3A+24-16A=5, giving A=1.
A=1,\quad B=-5,\quad C=-1,\quad D=2
Check with the x coefficient: 3A+16C-8D=3-16-16=-29, which matches the numerator.
Write the integrand in partial fractions. Split the last fraction so that one part has the derivative of x^{2}+3 on top.
\dfrac{1}{x-4}-\dfrac{5}{(x-4)^{2}}+\dfrac{-x+2}{x^{2}+3}=\dfrac{1}{x-4}-\dfrac{5}{(x-4)^{2}}-\dfrac12\cdot\dfrac{2x}{x^{2}+3}+\dfrac{2}{x^{2}+3}
Integrate each term: \int\dfrac{dx}{(x-4)^{2}}=-\dfrac{1}{x-4}, \int\dfrac{2x}{x^{2}+3}dx=\ln(x^{2}+3) and \int\dfrac{dx}{x^{2}+a^{2}}=\dfrac1a\tan^{-1}\dfrac xa with a=\sqrt3.
\int(\dots)\,dx=\ln|x-4|+\dfrac{5}{x-4}-\dfrac12\ln(x^{2}+3)+\dfrac{2}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3}+c

\ln|x-4|+\dfrac{5}{x-4}-\dfrac12\ln(x^{2}+3)+\dfrac{2}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3}+c

(v) \displaystyle\int\sin^{-1}x\,dx

Write the integrand as (\sin^{-1}x)(1). Take \sin^{-1}x as the first function (inverse trigonometric) and 1 as the second function, and integrate by parts.
\int\sin^{-1}x\,dx=\sin^{-1}x\cdot x-\int\dfrac{1}{\sqrt{1-x^{2}}}\cdot x\,dx
In the new integral put u=1-x^{2}, so du=-2x\,dx.
\int\dfrac{x}{\sqrt{1-x^{2}}}dx=-\dfrac12\int u^{-\frac12}du=-u^{\frac12}=-\sqrt{1-x^{2}}
Combine. Subtracting -\sqrt{1-x^{2}} gives a plus sign.
\int\sin^{-1}x\,dx=x\sin^{-1}x+\sqrt{1-x^{2}}+c

x\sin^{-1}x+\sqrt{1-x^{2}}+c

(vi) \displaystyle\int2x\sin3x\,dx

Integrate by parts. Take 2x as the first function (its derivative is 2) and \sin3x as the second (its integral is -\tfrac13\cos3x).
\int2x\sin3x\,dx=2x\left(-\dfrac{\cos3x}{3}\right)-\int2\left(-\dfrac{\cos3x}{3}\right)dx
Simplify and integrate the remaining term.
=-\dfrac{2x\cos3x}{3}+\dfrac23\int\cos3x\,dx=-\dfrac{2x\cos3x}{3}+\dfrac23\cdot\dfrac{\sin3x}{3}+c
Final form.
=-\dfrac23x\cos3x+\dfrac29\sin3x+c

-\dfrac23x\cos3x+\dfrac29\sin3x+c

(vii) \displaystyle\int x^{2}e^{x}dx

Integrate by parts with x^{2} as the first function and e^{x} as the second.
\int x^{2}e^{x}dx=x^{2}e^{x}-\int2x\,e^{x}dx
The new integral still contains a power of x, so use parts again with x first and e^{x} second.
\int2x\,e^{x}dx=2\left(xe^{x}-\int e^{x}dx\right)=2xe^{x}-2e^{x}
Combine, taking care with the signs.
\int x^{2}e^{x}dx=x^{2}e^{x}-2xe^{x}+2e^{x}+c=e^{x}(x^{2}-2x+2)+c

e^{x}(x^{2}-2x+2)+c

(viii) \displaystyle\int_{0}^{\pi/4}(\sin2x-5\cos4x)\,dx

Use \int\sin ax\,dx=-\dfrac{\cos ax}{a} and \int\cos ax\,dx=\dfrac{\sin ax}{a}.
\int_{0}^{\pi/4}(\sin2x-5\cos4x)\,dx=\left[-\dfrac{\cos2x}{2}-\dfrac{5\sin4x}{4}\right]_{0}^{\pi/4}
Upper limit: \cos\tfrac\pi2=0 and \sin\pi=0. Lower limit: \cos0=1 and \sin0=0.
=\left(0-0\right)-\left(-\dfrac12-0\right)=\dfrac12

\dfrac12

(ix) \displaystyle\int_{1}^{4}\dfrac{\cos\sqrt x}{2\sqrt x}dx

Substitute u=\sqrt x, so du=\dfrac{dx}{2\sqrt x}. Change the limits: x=1 gives u=1 and x=4 gives u=2.
\int_{1}^{4}\dfrac{\cos\sqrt x}{2\sqrt x}dx=\int_{1}^{2}\cos u\,du=\left[\sin u\right]_{1}^{2}
Evaluate in radians.
=\sin2-\sin1\approx0.9093-0.8415=0.0678

This is the same integral as Exercise 3.7, Question 10.

\sin2-\sin1\approx0.068

Use the substitution u=2x+1 to evaluate \displaystyle\int_{0}^{1}\dfrac{x^{2}}{\sqrt{2x+1}}\,dx.

Solution

Express x, x^{2} and dx in terms of u.
u=2x+1\;\Longrightarrow\;x=\dfrac{u-1}{2},\quad dx=\dfrac{du}{2},\quad x^{2}=\dfrac{(u-1)^{2}}{4}
Change the limits.
x=0\Rightarrow u=1,\qquad x=1\Rightarrow u=3
Substitute everything.
\int_{0}^{1}\dfrac{x^{2}}{\sqrt{2x+1}}dx=\int_{1}^{3}\dfrac{(u-1)^{2}}{4\sqrt u}\cdot\dfrac{du}{2}=\dfrac18\int_{1}^{3}\dfrac{u^{2}-2u+1}{u^{\frac12}}\,du
Divide each term by u^{1/2} and integrate with the power rule.
=\dfrac18\int_{1}^{3}\left(u^{\frac32}-2u^{\frac12}+u^{-\frac12}\right)du=\dfrac18\left[\dfrac25u^{\frac52}-\dfrac43u^{\frac32}+2u^{\frac12}\right]_{1}^{3}
Value at u=3, using 3^{5/2}=9\sqrt3, 3^{3/2}=3\sqrt3 and 3^{1/2}=\sqrt3.
\dfrac25(9\sqrt3)-\dfrac43(3\sqrt3)+2\sqrt3=\dfrac{18\sqrt3}{5}-4\sqrt3+2\sqrt3=\dfrac{8\sqrt3}{5}
Value at u=1.
\dfrac25-\dfrac43+2=\dfrac{6-20+30}{15}=\dfrac{16}{15}
Subtract and multiply by \tfrac18.
\dfrac18\left(\dfrac{8\sqrt3}{5}-\dfrac{16}{15}\right)=\dfrac{\sqrt3}{5}-\dfrac{2}{15}=\dfrac{3\sqrt3-2}{15}\approx0.213

\int_{0}^{1}\dfrac{x^{2}}{\sqrt{2x+1}}\,dx=\dfrac{3\sqrt3-2}{15}\approx0.213

A model rocket is launched upward from ground level with an initial speed of 60\,\mathrm{m/s}. (a) How long does it take for the rocket to reach its highest point? (b) How high does the rocket go?

Solution

Take upwards as positive, ground level as S=0 and the launch at t=0. From the Key Facts for upward motion: S(0)=0, v(0)=60\,\mathrm{m\,s^{-1}} and a=-9.8\,\mathrm{m\,s^{-2}} (gravity acts downwards).

Integrate the acceleration to get the velocity, and use v(0)=60 to find the constant.
v(t)=\int(-9.8)\,dt=-9.8t+c_{1},\qquad v(0)=60\;\Longrightarrow\;c_{1}=60
Integrate the velocity to get the position, and use S(0)=0.
S(t)=\int(60-9.8t)\,dt=60t-4.9t^{2}+c_{2},\qquad S(0)=0\;\Longrightarrow\;c_{2}=0

(a) How long does it take for the rocket to reach its highest point?

At the highest point the rocket stops rising for an instant, so its velocity is zero.
60-9.8t=0\;\Longrightarrow\;t=\dfrac{60}{9.8}\approx6.12\,\mathrm{s}

t\approx6.12\,\mathrm{s}

(b) How high does the rocket go?

Put t=\dfrac{60}{9.8} into S(t).
S_{\max}=60\left(\dfrac{60}{9.8}\right)-4.9\left(\dfrac{60}{9.8}\right)^{2}=\dfrac{3600}{9.8}-\dfrac{3600}{19.6}=\dfrac{3600}{19.6}
Evaluate.
S_{\max}\approx183.7\,\mathrm{m}

Height of the rocket against time. The top of the parabola is at t\approx6.12\,\mathrm{s}, height \approx183.7\,\mathrm{m}.

S_{\max}\approx183.7\,\mathrm{m}

Note. The book's Key Facts box prints a=g=-98\,\mathrm{m/s^{2}}; this is a misprint for -9.8\,\mathrm{m\,s^{-2}}. With g=10\,\mathrm{m\,s^{-2}} the answers would be 6\,\mathrm{s} and 180\,\mathrm{m}.

Suppose that a parachute moves with a velocity V(t)=\cos\pi t\,\mathrm{m/s} along a coordinate line. Assuming that the parachute has the coordinate S=4\,\mathrm{m} at time t=0\,\mathrm{s}, find its position.

Solution

Position is an anti-derivative of velocity, S(t)=\int V(t)\,dt.
S(t)=\int\cos\pi t\,dt=\dfrac{\sin\pi t}{\pi}+c
Use the initial condition S(0)=4 to find c.
4=\dfrac{\sin0}{\pi}+c\;\Longrightarrow\;c=4
Write the position function.
S(t)=\dfrac{\sin\pi t}{\pi}+4\ \ \mathrm{m}
Check: S'(t)=\dfrac{\pi\cos\pi t}{\pi}=\cos\pi t=V(t), and S(0)=0+4=4.

The parachute moves back and forth about S=4\,\mathrm{m}.

S(t)=4+\dfrac{\sin\pi t}{\pi}\ \ \mathrm{m}

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 31 mark

If \int xe^{x}\,dx=e^{x}(x-1)+C, then \int_{0}^{1}xe^{x}\,dx equals:

A0
B1Answer
Ce-1
D1-e

Why

By the Fundamental Theorem, \int_{0}^{1}xe^{x}\,dx=\Big[e^{x}(x-1)\Big]_{0}^{1}=e^{1}(0)-e^{0}(-1)=0+1=1.

Option C, e-1, is the value of \int_{0}^{1}e^{x}\,dx; it ignores the factor x.

2026 Annual I, Set B, MCQ 111 mark

Volume generated by revolving the region under y=\sqrt{x} from x=0 to x=4 about the x-axis is:

A4\pi
B6\pi
C8\piAnswer
D10\pi

Why

Disc method: V=\pi\int_{0}^{4}(\sqrt{x})^{2}\,dx=\pi\int_{0}^{4}x\,dx=\pi\left[\dfrac{x^{2}}{2}\right]_{0}^{4}=\pi\cdot\dfrac{16}{2}=8\pi.

Forgetting to square \sqrt{x} gives \tfrac{16}{3}\pi, and forgetting the \tfrac12 gives 16\pi; neither is an option.

2026 Annual I, Set B, MCQ 201 mark

Average value of f(x)=x^{2} on [0,3] is:

A6
B5
C4
D3Answer

Why

f_{\text{avg}}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx=\dfrac{1}{3-0}\int_{0}^{3}x^{2}\,dx=\dfrac13\left[\dfrac{x^{3}}{3}\right]_{0}^{3}=\dfrac13(9)=3.

The integral alone is 9; remember to divide by the length of the interval, b-a=3.

2026 Annual I, Set B, Q.2 (iii)4 marks

Evaluate \int\sin^{2}x\,dx

Solution

\sin^{2}x has no direct formula. Use the double-angle identity \cos2x=1-2\sin^{2}x to lower the power.
\sin^{2}x=\dfrac{1-\cos2x}{2}
Substitute and split into two integrals.
\int\sin^{2}x\,dx=\int\dfrac{1-\cos2x}{2}\,dx=\dfrac12\int dx-\dfrac12\int\cos2x\,dx
Integrate each part, using \int\cos2x\,dx=\dfrac{\sin2x}{2}.
=\dfrac12x-\dfrac12\cdot\dfrac{\sin2x}{2}+c=\dfrac{x}{2}-\dfrac{\sin2x}{4}+c
Check by differentiating.
\dfrac{d}{dx}\left(\dfrac{x}{2}-\dfrac{\sin2x}{4}\right)=\dfrac12-\dfrac{\cos2x}{2}=\dfrac{1-\cos2x}{2}=\sin^{2}x\ \checkmark

\dfrac{x}{2}-\dfrac{\sin 2x}{4}+c

Note. Marking: the identity, splitting the integral, and each integrated term earn the marks. The same method is Example 6 of the book.

2026 Annual I, Set B, Q.2 (ix) OR4 marks

Evaluate \int_{0}^{1}x\,e^{x}\,dx using integration by parts.

Solution

Take x as the first function (its derivative, 1, is simpler) and e^{x} as the second function (easy to integrate).
\int xe^{x}\,dx=x\int e^{x}\,dx-\int\left[\dfrac{d}{dx}(x)\int e^{x}\,dx\right]dx
Simplify and integrate the remaining integral.
=xe^{x}-\int1\cdot e^{x}\,dx=xe^{x}-e^{x}=e^{x}(x-1)
Apply the limits: value at the upper limit minus value at the lower limit.
\int_{0}^{1}xe^{x}\,dx=\Big[e^{x}(x-1)\Big]_{0}^{1}=e^{1}(1-1)-e^{0}(0-1)=0-(-1)=1

\int_{0}^{1}xe^{x}\,dx=1

2026 Annual I, Set B, Q.2 (x) OR4 marks

Find the area bounded by the curve y=x^{2}+2, the x-axis, and the lines x=1 and x=3.

Solution

y=x^{2}+2\ge2>0 for every x, so the curve lies above the x-axis on [1,3] and the area is the integral itself.
A=\int_{1}^{3}(x^{2}+2)\,dx
Integrate term by term.
=\left[\dfrac{x^{3}}{3}+2x\right]_{1}^{3}
Apply the limits.
=\left(\dfrac{27}{3}+6\right)-\left(\dfrac13+2\right)=15-\dfrac73=\dfrac{38}{3}

The shaded region lies under y=x^{2}+2, above the x-axis, between x=1 and x=3.

A=\dfrac{38}{3}\approx12.67\ \text{sq. units}

2026 Annual I, Set B, Q.5 OR8 marks

(a) Resolve into partial fractions: \dfrac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)} (b) Hence find: \displaystyle\int\frac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)}\,dx

Solution

(a) Resolve into partial fractions.

The denominator has a linear factor x-1 and a quadratic factor x^{2}+2x+5 with no real factors (its discriminant is 4-20<0). The degree of the numerator (2) is less than that of the denominator (3), so the fraction is proper.
\dfrac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+2x+5}
Multiply both sides by (x^{2}+2x+5)(x-1).
x^{2}+3x+4=A(x^{2}+2x+5)+(Bx+C)(x-1)
Put x=1; the second term vanishes.
1+3+4=A(1+2+5)\;\Longrightarrow\;8=8A\;\Longrightarrow\;A=1
Compare the coefficients of x^{2} and the constant terms.
x^{2}:\ 1=A+B\;\Rightarrow\;B=0;\qquad\text{constant}:\ 4=5A-C\;\Rightarrow\;C=1
Check with the coefficient of x: 2A-B+C=2-0+1=3, which matches. So:
\dfrac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)}=\dfrac{1}{x-1}+\dfrac{1}{x^{2}+2x+5}

\dfrac{1}{x-1}+\dfrac{1}{x^{2}+2x+5}

(b) Hence find the integral.

Integrate the two partial fractions separately.
\int\dfrac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)}\,dx=\int\dfrac{dx}{x-1}+\int\dfrac{dx}{x^{2}+2x+5}
The first gives \ln|x-1|. For the second, complete the square, x^{2}+2x+5=(x+1)^{2}+2^{2}, and use \int\dfrac{du}{u^{2}+a^{2}}=\dfrac1a\tan^{-1}\dfrac{u}{a} with u=x+1, a=2.
\int\dfrac{dx}{(x+1)^{2}+2^{2}}=\dfrac12\tan^{-1}\left(\dfrac{x+1}{2}\right)
Combine the results, with one constant.
\int\dfrac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)}\,dx=\ln|x-1|+\dfrac12\tan^{-1}\left(\dfrac{x+1}{2}\right)+c

\ln|x-1|+\dfrac{1}{2}\tan^{-1}\left(\dfrac{x+1}{2}\right)+c

Note. No split between (a) and (b) is printed. Expect roughly half the marks for the partial fractions and half for the integration.

2026 Annual I, Set D, MCQ 11 mark

Average value of f(x)=x^{2} on [1,3] is:

A\dfrac{28}{3}
B\dfrac{26}{3}
C\dfrac{13}{3}Answer
D\dfrac{14}{3}

Why

f_{\text{avg}}=\dfrac{1}{3-1}\int_{1}^{3}x^{2}\,dx=\dfrac12\left[\dfrac{x^{3}}{3}\right]_{1}^{3}=\dfrac12\left(9-\dfrac13\right)=\dfrac12\cdot\dfrac{26}{3}=\dfrac{13}{3}.

Option B, \tfrac{26}{3}, is the integral itself; it has not been divided by b-a=2.

2026 Annual I, Set D, MCQ 41 mark

If \displaystyle\int\frac{dx}{\sqrt{4-x^{2}}}=\sin^{-1}\left(\frac{x}{2}\right)+C, then \displaystyle\int_{0}^{1}\frac{dx}{\sqrt{4-x^{2}}} equals:

A\pi
B\dfrac{\pi}{6}Answer
C\dfrac{\pi}{3}
D\dfrac{\pi}{2}

Why

\Big[\sin^{-1}\dfrac{x}{2}\Big]_{0}^{1}=\sin^{-1}\dfrac12-\sin^{-1}0=\dfrac{\pi}{6}-0=\dfrac{\pi}{6}.

Option C, \tfrac{\pi}{3}, would come from \cos^{-1}\tfrac12; the inverse sine of \tfrac12 is \tfrac{\pi}{6}.

2026 Annual I, Set D, MCQ 121 mark

The volume generated by revolving the region under y=\sqrt{x} from x=0 to x=16 about the x-axis is:

A256\pi
B32\pi
C64\pi
D128\piAnswer

Why

V=\pi\int_{0}^{16}(\sqrt{x})^{2}\,dx=\pi\int_{0}^{16}x\,dx=\pi\left[\dfrac{x^{2}}{2}\right]_{0}^{16}=\pi\cdot\dfrac{256}{2}=128\pi.

Option A, 256\pi, forgets the \tfrac12 from \int x\,dx=\tfrac{x^{2}}{2}.

2026 Annual I, Set D, Q.2 (iii)4 marks

Evaluate the integral: \int\cos^{2}(2\theta)\,d\theta

Solution

Lower the power with \cos^{2}A=\dfrac{1+\cos2A}{2}, taking A=2\theta, so that 2A=4\theta.
\cos^{2}(2\theta)=\dfrac{1+\cos4\theta}{2}
Substitute and split into two integrals.
\int\cos^{2}(2\theta)\,d\theta=\dfrac12\int d\theta+\dfrac12\int\cos4\theta\,d\theta
Integrate each part, using \int\cos4\theta\,d\theta=\dfrac{\sin4\theta}{4}.
=\dfrac{\theta}{2}+\dfrac12\cdot\dfrac{\sin4\theta}{4}+c=\dfrac{\theta}{2}+\dfrac{\sin4\theta}{8}+c
Check by differentiating.
\dfrac{d}{d\theta}\left(\dfrac{\theta}{2}+\dfrac{\sin4\theta}{8}\right)=\dfrac12+\dfrac{\cos4\theta}{2}=\dfrac{1+\cos4\theta}{2}=\cos^{2}(2\theta)\ \checkmark

\dfrac{\theta}{2}+\dfrac{\sin 4\theta}{8}+c

2026 Annual I, Set D, Q.2 (ix) OR4 marks

Use the method of integration by parts to evaluate: \int_{1}^{2}x(\ln x)\,dx

Solution

Take \ln x as the first function (easy to differentiate, hard to integrate) and x as the second function.
\int x\ln x\,dx=\ln x\int x\,dx-\int\left[\dfrac{d}{dx}(\ln x)\int x\,dx\right]dx=\dfrac{x^{2}}{2}\ln x-\int\dfrac1x\cdot\dfrac{x^{2}}{2}\,dx
Simplify the remaining integral and integrate.
=\dfrac{x^{2}}{2}\ln x-\dfrac12\int x\,dx=\dfrac{x^{2}}{2}\ln x-\dfrac{x^{2}}{4}
Apply the limits, using \ln1=0.
\int_{1}^{2}x\ln x\,dx=\left[\dfrac{x^{2}}{2}\ln x-\dfrac{x^{2}}{4}\right]_{1}^{2}=\left(2\ln2-1\right)-\left(0-\dfrac14\right)=2\ln2-\dfrac34
As a decimal, with \ln2\approx0.6931:
2(0.6931)-0.75\approx0.636

\int_{1}^{2}x\ln x\,dx=2\ln 2-\dfrac{3}{4}\approx0.636

2026 Annual I, Set D, Q.2 (x) OR4 marks

Find area of the region bounded by the curves y=x^{2} and y=4x-x^{2}.

Solution

Find the intersection points of the two curves.
x^{2}=4x-x^{2}\;\Longrightarrow\;2x^{2}-4x=0\;\Longrightarrow\;2x(x-2)=0\;\Longrightarrow\;x=0,\ x=2
The points are (0,0) and (2,4). Test x=1: 4x-x^{2}=3 and x^{2}=1, so y=4x-x^{2} is the upper curve.
4x-x^{2}\ge x^{2}\quad\text{on}\quad[0,2]
Integrate upper minus lower between the intersection points.
A=\int_{0}^{2}\big[(4x-x^{2})-x^{2}\big]\,dx=\int_{0}^{2}(4x-2x^{2})\,dx=\left[2x^{2}-\dfrac{2x^{3}}{3}\right]_{0}^{2}
Apply the limits.
=\left(8-\dfrac{16}{3}\right)-0=\dfrac83

The region between the parabola y=x^{2} (below) and y=4x-x^{2} (above), from x=0 to x=2.

A=\dfrac{8}{3}\approx2.67\ \text{sq. units}

2026 Annual I, Set D, Q.5 OR8 marks

(a) Resolve into partial fractions: \dfrac{3x^{2}+2x+7}{(x^{2}+4x+5)(x-3)} (b) Hence find: \displaystyle\int\frac{3x^{2}+2x+7}{(x^{2}+4x+5)(x-3)}\,dx

Solution

(a) Resolve into partial fractions.

x-3 is a linear factor and x^{2}+4x+5 has no real factors (discriminant 16-20<0). The fraction is proper.
\dfrac{3x^{2}+2x+7}{(x^{2}+4x+5)(x-3)}=\dfrac{A}{x-3}+\dfrac{Bx+C}{x^{2}+4x+5}
Multiply both sides by (x^{2}+4x+5)(x-3).
3x^{2}+2x+7=A(x^{2}+4x+5)+(Bx+C)(x-3)
Put x=3.
27+6+7=A(9+12+5)\;\Longrightarrow\;40=26A\;\Longrightarrow\;A=\dfrac{20}{13}
Compare the coefficients of x^{2} and the constant terms.
\begin{aligned}x^{2}:\ 3&=A+B\\ \Rightarrow\ B&=3-\dfrac{20}{13}=\dfrac{19}{13}\\ \text{constant}:\ 7&=5A-3C\\ \Rightarrow\ 3C&=\dfrac{100}{13}-7=\dfrac{9}{13}\\ \Rightarrow\ C&=\dfrac{3}{13}\end{aligned}
Check with the coefficient of x: 4A-3B+C=\dfrac{80-57+3}{13}=\dfrac{26}{13}=2, which matches. So:
\dfrac{3x^{2}+2x+7}{(x^{2}+4x+5)(x-3)}=\dfrac{20}{13(x-3)}+\dfrac{19x+3}{13(x^{2}+4x+5)}

\dfrac{20}{13(x-3)}+\dfrac{19x+3}{13(x^{2}+4x+5)}

(b) Hence find the integral.

Integrate the linear part.
\int\dfrac{20}{13(x-3)}\,dx=\dfrac{20}{13}\ln|x-3|
For the quadratic part, write the numerator as a multiple of the derivative 2x+4 plus a constant: 19x+3=\tfrac{19}{2}(2x+4)-35.
\dfrac{19x+3}{x^{2}+4x+5}=\dfrac{19}{2}\cdot\dfrac{2x+4}{x^{2}+4x+5}-\dfrac{35}{(x+2)^{2}+1}
The first piece is 'derivative over function' and gives a logarithm; the second gives an inverse tangent, since x^{2}+4x+5=(x+2)^{2}+1.
\int\dfrac{19x+3}{x^{2}+4x+5}\,dx=\dfrac{19}{2}\ln(x^{2}+4x+5)-35\tan^{-1}(x+2)
Multiply by \tfrac{1}{13} and add the linear part.
\int\dfrac{3x^{2}+2x+7}{(x^{2}+4x+5)(x-3)}\,dx=\dfrac{20}{13}\ln|x-3|+\dfrac{19}{26}\ln(x^{2}+4x+5)-\dfrac{35}{13}\tan^{-1}(x+2)+c

\dfrac{20}{13}\ln|x-3|+\dfrac{19}{26}\ln(x^{2}+4x+5)-\dfrac{35}{13}\tan^{-1}(x+2)+c

Note. The constants are fractions (A=\tfrac{20}{13}, B=\tfrac{19}{13}, C=\tfrac{3}{13}); this is how the printed question works out, not a slip. No split between (a) and (b) is printed.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 3, old syllabus1 mark

\displaystyle\int\frac{1}{x}\,dx=

A\ln|x|+CAnswer
Be^{x}+C
Cx\ln x+C
D\dfrac{x^{2}}{2}+C

Why

The derivative of \ln|x| is \dfrac1x (Formula 3.2), so \int\dfrac{dx}{x}=\ln|x|+C. The power rule cannot be used because n=-1.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 10, old syllabus1 mark

What results \int_{0}^{1}x^{2}\,dx ?

A\dfrac13Answer
B1
C\dfrac12
D0

Why

\int_{0}^{1}x^{2}\,dx=\left[\dfrac{x^{3}}{3}\right]_{0}^{1}=\dfrac13-0=\dfrac13.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 19, old syllabus1 mark

To evaluate \int2xe^{x^{2}}\,dx, the appropriate substitution is:

Au=x^{2}Answer
Bu=e^{x^{2}}
Cu=2x
Du=x

Why

With u=x^{2}, du=2x\,dx, so \int2xe^{x^{2}}dx=\int e^{u}\,du=e^{u}+C=e^{x^{2}}+C. The factor 2x is exactly the derivative of the power x^{2}.

Option B also happens to work (du=2xe^{x^{2}}dx turns the integral into \int du), but the standard substitution the question expects is the power of e, u=x^{2}.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (iii), old syllabus4 marks

Evaluate the integral using substitution x=5\tan\theta: \displaystyle\int\frac{1}{\sqrt{x^{2}+25}}\,dx

Solution

Substitute x=5\tan\theta with -\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2}, so that \sec\theta>0.
x=5\tan\theta\;\Longrightarrow\;dx=5\sec^{2}\theta\,d\theta
Simplify the root with 1+\tan^{2}\theta=\sec^{2}\theta.
\sqrt{x^{2}+25}=\sqrt{25\tan^{2}\theta+25}=\sqrt{25\sec^{2}\theta}=5\sec\theta
Substitute and cancel; then use \int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|.
\int\dfrac{dx}{\sqrt{x^{2}+25}}=\int\dfrac{5\sec^{2}\theta}{5\sec\theta}\,d\theta=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+c
Return to x: \tan\theta=\dfrac{x}{5} and \sec\theta=\sqrt{1+\tan^{2}\theta}=\dfrac{\sqrt{x^{2}+25}}{5}.
=\ln\left|\dfrac{x+\sqrt{x^{2}+25}}{5}\right|+c=\ln\left(x+\sqrt{x^{2}+25}\right)-\ln5+c
c-\ln5 is another constant, C. No absolute value is needed, because x+\sqrt{x^{2}+25}>0.
\int\dfrac{dx}{\sqrt{x^{2}+25}}=\ln\left(x+\sqrt{x^{2}+25}\right)+C

\ln\left(x+\sqrt{x^{2}+25}\right)+C

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (viii) OR, old syllabus4 marks

Find the total area bounded by y=x^{3}-4x, the x-axis, and the vertical lines x=-2 and x=2.

Solution

Find where the curve meets the x-axis.
x^{3}-4x=x(x-2)(x+2)=0\;\Longrightarrow\;x=-2,\ 0,\ 2
Check the signs. At x=-1, y=-1+4=3>0, so the curve is above the axis on (-2,0). At x=1, y=1-4=-3<0, so it is below the axis on (0,2). Split the area at x=0.
Integrate over [-2,0].
\int_{-2}^{0}(x^{3}-4x)\,dx=\left[\dfrac{x^{4}}{4}-2x^{2}\right]_{-2}^{0}=0-(4-8)=4
Integrate over [0,2]. The negative value shows that this part is below the axis.
\int_{0}^{2}(x^{3}-4x)\,dx=\left[\dfrac{x^{4}}{4}-2x^{2}\right]_{0}^{2}=(4-8)-0=-4
Total area: add the absolute values.
A=4+|-4|=8

The curve is above the axis on (-2,0) and below it on (0,2). Both regions have area 4.

A=8\ \text{sq. units}

2026 Annual I, Old syllabus (Curriculum 2000), Q.5, old syllabus8 marks

Evaluate using the method of partial fractions and express your answer as a single logarithm: \displaystyle\int\frac{6x^{2}+11x+3}{(x+1)(x+2)(x+3)}\,dx

Solution

The numerator has degree 2 and the denominator degree 3, so the fraction is proper. The three factors are distinct and linear.
\dfrac{6x^{2}+11x+3}{(x+1)(x+2)(x+3)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{x+3}
Multiply by (x+1)(x+2)(x+3).
6x^{2}+11x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)
Put x=-1: 6-11+3=A(1)(2).
-2=2A\;\Longrightarrow\;A=-1
Put x=-2: 24-22+3=B(-1)(1).
5=-B\;\Longrightarrow\;B=-5
Put x=-3: 54-33+3=C(-2)(-1).
24=2C\;\Longrightarrow\;C=12
Check: comparing the x^{2} coefficients gives A+B+C=6, and indeed -1-5+12=6. Now integrate each term.
\int\left[-\dfrac{1}{x+1}-\dfrac{5}{x+2}+\dfrac{12}{x+3}\right]dx=-\ln|x+1|-5\ln|x+2|+12\ln|x+3|+c
Combine into one logarithm with k\ln a=\ln a^{k}, \ln a+\ln b=\ln ab and \ln a-\ln b=\ln\dfrac{a}{b}.
=\ln|x+3|^{12}-\ln|x+1|-\ln|x+2|^{5}+c=\ln\left|\dfrac{(x+3)^{12}}{(x+1)(x+2)^{5}}\right|+c

\ln\left|\dfrac{(x+3)^{12}}{(x+1)(x+2)^{5}}\right|+c

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 8, old syllabus1 mark

The evaluated value of \int xe^{x^{2}}\,dx is:

Ae^{x^{2}}+c
B\dfrac12e^{x^{2}}+cAnswer
C2e^{x^{2}}+c
Dx^{2}e^{x^{2}}+c

Why

Put t=x^{2}, so x\,dx=\tfrac12dt: \int xe^{x^{2}}dx=\tfrac12\int e^{t}dt=\tfrac12e^{x^{2}}+c. Check: \dfrac{d}{dx}\left(\tfrac12e^{x^{2}}\right)=\tfrac12e^{x^{2}}(2x)=xe^{x^{2}}.

Option A differentiates to 2xe^{x^{2}}, twice the integrand.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 9, old syllabus1 mark

The evaluated value of \int_{-2}^{2}(x^{4}+1)\,dx

A0
B2\int_{-2}^{2}(x^{4})\,dx
C\int_{-2}^{2}(x^{4}+1)\,dx
D2\int_{-2}^{2}(x^{4}+1)\,dxAnswer

Why

x^{4}+1 is an even function, so \int_{-2}^{2}(x^{4}+1)\,dx=2\int_{0}^{2}(x^{4}+1)\,dx=2\left[\dfrac{x^{5}}{5}+x\right]_{0}^{2}=2\left(\dfrac{32}{5}+2\right)=\dfrac{84}{5}. This is option D with the lower limit 0.

Option A (0) would be the answer for an odd function. Option B leaves out the +1.

Note. Paper misprint: every option is printed with the lower limit -2. The intended option D is 2\int_{0}^{2}(x^{4}+1)\,dx; as printed, D would be double the value, and option C merely repeats the question, so C is the only printed option that is literally equal to it.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i) OR, old syllabus4 marks

Evaluate \displaystyle\int\frac{dx}{x^{2}\sqrt{x^{2}-9}}

Solution

The root \sqrt{x^{2}-9}=\sqrt{x^{2}-3^{2}} suggests x=3\sec\theta. Take 0<\theta<\tfrac{\pi}{2} (that is, x>3), so \tan\theta>0.
x=3\sec\theta\;\Longrightarrow\;dx=3\sec\theta\tan\theta\,d\theta
Simplify the root and x^{2} using \sec^{2}\theta-1=\tan^{2}\theta.
\sqrt{x^{2}-9}=\sqrt{9\sec^{2}\theta-9}=3\tan\theta,\qquad x^{2}=9\sec^{2}\theta
Substitute and cancel.
\int\dfrac{3\sec\theta\tan\theta\,d\theta}{9\sec^{2}\theta\cdot3\tan\theta}=\dfrac19\int\dfrac{d\theta}{\sec\theta}=\dfrac19\int\cos\theta\,d\theta=\dfrac19\sin\theta+c
Return to x. From \sec\theta=\dfrac{x}{3}, draw a right triangle with hypotenuse x, adjacent side 3 and opposite side \sqrt{x^{2}-9}.
\sin\theta=\dfrac{\sqrt{x^{2}-9}}{x}
Substitute.
\int\dfrac{dx}{x^{2}\sqrt{x^{2}-9}}=\dfrac{\sqrt{x^{2}-9}}{9x}+c
Check by differentiating with the quotient rule.
\dfrac{d}{dx}\left[\dfrac{\sqrt{x^{2}-9}}{9x}\right]=\dfrac19\cdot\dfrac{x\cdot\frac{x}{\sqrt{x^{2}-9}}-\sqrt{x^{2}-9}}{x^{2}}=\dfrac19\cdot\dfrac{x^{2}-(x^{2}-9)}{x^{2}\sqrt{x^{2}-9}}=\dfrac{1}{x^{2}\sqrt{x^{2}-9}}\ \checkmark

\dfrac{\sqrt{x^{2}-9}}{9x}+c

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii) OR, old syllabus4 marks

Find area between y=x^{2}-3x and x-axis from x=0 to x=4

Solution

The curve meets the x-axis where x^{2}-3x=x(x-3)=0, at x=0 and x=3. It is below the axis on (0,3) (at x=1, y=-2) and above it on (3,4).
x(x-3)=0\;\Longrightarrow\;x=0,\ x=3
Integrate over [0,3].
\int_{0}^{3}(x^{2}-3x)\,dx=\left[\dfrac{x^{3}}{3}-\dfrac{3x^{2}}{2}\right]_{0}^{3}=9-\dfrac{27}{2}=-\dfrac92
Integrate over [3,4].
\int_{3}^{4}(x^{2}-3x)\,dx=\left(\dfrac{64}{3}-24\right)-\left(9-\dfrac{27}{2}\right)=-\dfrac83+\dfrac92=\dfrac{11}{6}
Add the absolute values.
A=\dfrac92+\dfrac{11}{6}=\dfrac{27+11}{6}=\dfrac{38}{6}=\dfrac{19}{3}

A_{1}=\tfrac92 lies below the axis on [0,3] and A_{2}=\tfrac{11}{6} lies above it on [3,4].

A=\dfrac{19}{3}\approx6.33\ \text{sq. units}

2026 Annual I, Old syllabus (Curriculum 2006), Q.5 OR, old syllabus8 marks

Using partial fraction, evaluate \displaystyle\int\frac{6x^{2}+5x+4}{x(x+1)(x-1)(x+2)}\,dx

Solution

The numerator has degree 2 and the denominator degree 4, so the fraction is proper. All four factors are distinct and linear.
\dfrac{6x^{2}+5x+4}{x(x+1)(x-1)(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{x-1}+\dfrac{D}{x+2}
Multiply by the denominator.
6x^{2}+5x+4=A(x+1)(x-1)(x+2)+Bx(x-1)(x+2)+Cx(x+1)(x+2)+Dx(x+1)(x-1)
Put x=0: 4=A(1)(-1)(2).
A=-2
Put x=-1: 6-5+4=B(-1)(-2)(1).
5=2B\;\Longrightarrow\;B=\dfrac52
Put x=1: 6+5+4=C(1)(2)(3).
15=6C\;\Longrightarrow\;C=\dfrac52
Put x=-2: 24-10+4=D(-2)(-1)(-3).
18=-6D\;\Longrightarrow\;D=-3
Check: the numerator has no x^{3} term, so A+B+C+D must be 0, and -2+\tfrac52+\tfrac52-3=0. Now integrate each term.
\int\left[-\dfrac{2}{x}+\dfrac{5/2}{x+1}+\dfrac{5/2}{x-1}-\dfrac{3}{x+2}\right]dx=-2\ln|x|+\dfrac52\ln|x+1|+\dfrac52\ln|x-1|-3\ln|x+2|+c
The answer may also be written as one logarithm.
=\ln\left|\dfrac{(x^{2}-1)^{5/2}}{x^{2}(x+2)^{3}}\right|+c

-2\ln|x|+\dfrac{5}{2}\ln|x+1|+\dfrac{5}{2}\ln|x-1|-3\ln|x+2|+c

Exam focus

Prove \int\sec x\,dx=\ln|\sec x+\tan x|+c; also the integrals of \operatorname{cosec} x, \tan x and \cot x. (short)
Standard 'prove that' results. Each proof is three lines using \int\dfrac{f'}{f}\,dx=\ln|f|.
Go to answer
Evaluate \int\sin^{2}x\,dx or \int\cos^{2}(kx)\,dx with the double-angle identities. (problem)
Asked in both 2026 new-syllabus sets (Q.2 (iii), 4 marks).
Go to answer
Derive Formulas 3.5 to 3.9 by trigonometric substitution. (long)
Book derivations; the 2026 old-syllabus papers asked \int\dfrac{dx}{\sqrt{x^{2}+25}} and \int\dfrac{dx}{x^{2}\sqrt{x^{2}-9}}.
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Evaluate \int_{0}^{1}xe^{x}\,dx and \int_{1}^{2}x\ln x\,dx by parts. (problem)
Q.2 (ix) OR in 2026 Sets B and D. Know how to choose the first function.
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Prove \int\sqrt{a^{2}-x^{2}}\,dx=\dfrac{x\sqrt{a^{2}-x^{2}}}{2}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a}+c by parts; evaluate \int e^{ax}\sin bx\,dx. (long)
Book examples in which the integral comes back; favourite long questions.
Go to answer
Resolve into partial fractions and integrate, for example \int\dfrac{x^{2}+3x+4}{(x^{2}+2x+5)(x-1)}\,dx. (long)
Q.5 OR (8 marks) in both 2026 sets, and Q.5 in both old-syllabus papers.
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Find an area under a curve, across the x-axis, or between two curves, with a shaded sketch. (problem)
Four area questions in the 2026 papers (Q.2 (x) OR in Sets B and D, and the old-syllabus papers).
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Find a volume of revolution with V=\pi\int_{a}^{b}y^{2}\,dx. (mcq)
2026 Set B MCQ 11 and Set D MCQ 12.
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Find the average value of a function on an interval. (mcq)
2026 Set B MCQ 20 and Set D MCQ 1; not in the book text but quick marks.
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Evaluate definite integrals with the Fundamental Theorem, including substitution with changed limits. (problem)
Many MCQs test \Big[F(x)\Big]_{a}^{b} and the properties; Example 25 shows both substitution methods.
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Use the general forms \int f^{n}f'\,dx, \int\dfrac{f'}{f}\,dx, \int e^{f}f'\,dx, for example \int xe^{x^{2}}\,dx. (mcq)
Three old-syllabus MCQs in 2026 and most of Exercise 3.1 use them.
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Consumer and producer surplus; work done in stretching a spring. (problem)
Application questions from Section 3.12 that follow a fixed pattern.
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