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Chapter 7: Integrals

35 words · 1 min read

Chapter 7

Integrals

Big idea: integration does two jobs that look different but are the same. It undoes differentiation, and it adds up infinitely many tiny pieces to find a total, such as an area.

7.1 Introduction

334 words · 2 min read

A tank is filling with water. A sensor tells you the rate at each moment: after $t$ minutes, water flows in at $3t^2$ litres per minute. The rate grows, so the tank fills faster and faster. How much water is in the tank after $4$ minutes?

In Chapter 6 you went from a quantity to its rate by differentiating. Now you must go backwards: from the rate to the quantity. You need a function $V(t)$ whose derivative is $3t^2$ . Since $\frac{d}{dt}t^3 = 3t^2$ , the volume is $V(t) = t^3$ , and after $4$ minutes there are $4^3 = 64$ litres.

A function that you differentiate to get $f$ is called an antiderivative of $f$ . Finding it is called integration.

There is a second way to look at the tank. Draw the rate $y = 3t^2$ against time. A small strip of width $\Delta t$ under the curve has area about (rate) $\times$ (time), which is the water that flows in during that short moment. Adding all the strips from $t = 0$ to $t = 4$ gives the total water, which is the area under the curve. The shaded area below is exactly $64$ square units, the same $64$ .

Graph of the curve y equals 3 t squared for t from 0 to 4, rising from 0 to 48. The region between the curve and the horizontal axis is shaded. Its area is 64.

The area under y = 3t² from t = 0 to t = 4 is 64. It equals the antiderivative t³ evaluated at 4.

This is the big discovery of the chapter: area under a curve and undoing a derivative are the same thing. In Sections 7.7 and 7.8 you will see why.

In this chapter you will learn: how to find antiderivatives (indefinite integrals) with substitution, trigonometric identities, partial fractions and integration by parts; and how to use them to calculate areas and totals (definite integrals) with the Fundamental Theorem of Calculus.

7.2 Integration as an Inverse Process of Differentiation

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Differentiation takes a function $F$ and gives $F'$ . Integration goes the other way: given $f$ , find a function $F$ with $F'(x) = f(x)$ . We write

$\int f(x) \, dx = F(x) + C \quad \text{when} \quad F'(x) = f(x).$

The sign $\int$ is the integral sign (a long "S" for "sum"). The function $f(x)$ is the integrand. The $dx$ says the variable is $x$ .
The function $F$ is an antiderivative of $f$ .
The number $C$ is the constant of integration. We will see in a moment why it must be there.

The derivative of $x^3$ is $3x^2$ . But the derivative of $x^3 + 5$ is also $3x^2$ , and so is the derivative of $x^3 - 2$ . A constant disappears when you differentiate. So when you go backwards you cannot know which constant was there, and you write all the possibilities at once:

$\int 3x^2 \, dx = x^3 + C.$

Geometrically, each value of $C$ gives one curve, and the curves are copies of each other shifted up or down. They all have the same slope above any given value of $x$ .

Three copies of the parabola y equals x squared plus C, for C equal to minus 2, 0 and 2, shifted vertically. A dashed vertical line at x equals 1 passes through the three curves at the points (1, minus 1), (1, 1) and (1, 3). The curves have the same slope at those points.

The antiderivatives x² + C for C = −2, 0, 2. Above x = 1 each curve has slope 2.

A point decides the constant. If the curve must pass through a known point, you can find $C$ . Suppose a curve has slope $2x$ at every point and passes through $(1, 3)$ . Then $y = \int 2x \, dx = x^2 + C$ . Putting in the point: $3 = 1 + C$ , so $C = 2$ , and the curve is $y = x^2 + 2$ .

Every derivative you know gives an integral for free, by reading it backwards. Each row of this table was checked by differentiating the answer.

Integrand $f(x)$	Antiderivative $F(x)$ (add $+C$ )
$x^n \; (n \ne -1)$	$\dfrac{x^{n+1}}{n + 1}$
$\dfrac{1}{x}$	$\ln\|x\|$
$e^x$	$e^x$
$a^x$	$\dfrac{a^x}{\ln a}$
$\sin x$	$-\cos x$
$\cos x$	$\sin x$
$\sec^2 x$	$\tan x$
$\csc^2 x$	$-\cot x$
$\sec x \tan x$	$\sec x$
$\csc x \cot x$	$-\csc x$
$\dfrac{1}{\sqrt{1 - x^2}}$	$\sin^{-1} x$
$\dfrac{1}{1 + x^2}$	$\tan^{-1} x$

In the first row, $n$ can be any number except $-1$ (for $n = -1$ use the second row). For $a^x$ , take $a > 0$ , $a \ne 1$ .

Always check by differentiating. Your answer

F

is right if

F'(x)

gives back the integrand. This takes a few seconds and catches most mistakes.

A cyclist starts at position $0$ and moves with speed $v(t) = 6t + 2$ metres per second. Her position $s(t)$ has derivative $v$ , so $s(t) = \int (6t + 2)\,dt = 3t^2 + 2t + C$ . At $t = 0$ she is at $0$ , so $C = 0$ . After $5$ seconds she has gone $s(5) = 3 \cdot 25 + 10 = 85$ metres.

7.2.1 Some properties of indefinite integral

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These rules let you break a hard integral into easy pieces. They come directly from the matching rules for derivatives.

Differentiating an integral gives back the function: $\dfrac{d}{dx}\int f(x)\,dx = f(x)$ .
Integrating a derivative gives back the function (plus a constant): $\int F'(x)\,dx = F(x) + C$ .
Constant multiple: $\int k\,f(x)\,dx = k\int f(x)\,dx$ .
Sum and difference: $\int \big(f(x) \pm g(x)\big)\,dx = \int f(x)\,dx \pm \int g(x)\,dx$ .

Together, the last two say that integration is linear: $\int \big(a f + b g\big)\,dx = a\int f\,dx + b\int g\,dx$ . Because all the pieces together give only one unknown constant, you write a single $+C$ at the end.

$\int (4x^3 - 6x^2 + 5)\,dx = x^4 - 2x^3 + 5x + C$

Integrate each term with the power rule: raise the power by $1$ , divide by the new power.

$\int \left(3\sin x + 2e^x - \frac{1}{x}\right) dx = -3\cos x + 2e^x - \ln|x| + C$

Find $\int \dfrac{(x^2 + 1)^2}{x}\,dx$ . There is no rule for a bracket divided by $x$ . So expand, then divide each term: $\dfrac{x^4 + 2x^2 + 1}{x} = x^3 + 2x + \dfrac{1}{x}$ . Then

$\int \left(x^3 + 2x + \frac{1}{x}\right) dx = \frac{x^4}{4} + x^2 + \ln|x| + C.$

Find $\int \sqrt{x}\,(x + 2)\,dx$ . Write $\sqrt{x} = x^{1/2}$ and expand: $x^{3/2} + 2x^{1/2}$ . Then

$\int \left(x^{3/2} + 2x^{1/2}\right) dx = \frac{2}{5}x^{5/2} + \frac{4}{3}x^{3/2} + C.$

Find $\int \tan^2 x\,dx$ . There is no rule for $\tan^2 x$ , but $\tan^2 x = \sec^2 x - 1$ . So $\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + C$ .

Common mistake: the integral of a product is not the product of the integrals. Compare

\int x \cdot x\,dx = \frac{x^3}{3} + C

with

\left(\int x\,dx\right)\left(\int x\,dx\right) = \frac{x^4}{4}

. The derivative of

\frac{x^4}{4}

is

x^3

, which is not

x^2

: at

x = 2

these are

8

and

4

.

Common mistake: forgetting

+ C

. Without it, you have named only one antiderivative, not all of them.

Find $\int 6x^2 - 4x + 1 \, dx$ .
Find $\int e^x + 4\cos x \, dx$ .
Find $\int \frac{x + 2}{x} \, dx$ .
Find $\int (x + 1)^2 \, dx$ .

Answers

$2 x^{3} - 2 x^{2} + x + C$
$e^{x} + 4 \sin x + C$
$x + 2\ln|x| + C$ (split into $1 + \frac{2}{x}$ first).
$\frac{x^3}{3} + x^2 + x + C$ (expand first).

EXERCISE 7.1

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Find $\int 3x^2 + 4x - 5 \, dx$ .
Find $\int \frac{5}{x^3} - 2\sqrt{x} \, dx$ .
Find $\int e^x + 3\cos x \, dx$ .
Find $\int (1 + x)^2 \, dx$ by expanding the bracket.
Find $\int \frac{x^3 + 2}{x^2} \, dx$ . (Divide each term first.)
Find $\int \sin^2\left(\frac{x}{2}\right) \, dx$ . (Use $\sin^2 A = \frac{1 - \cos 2A}{2}$ .)
Find $\int \frac{1}{1 + x^2} + \frac{2}{\sqrt{1 - x^2}} \, dx$ .
Find $\int 2^x + \frac{1}{x} \, dx$ .
Find $\int \tan^2 x + 1 \, dx$ .
A curve has slope $3x^2 - 2x$ at every point and passes through $(0, 1)$ . Find its equation. Where is it at $x = 2$ ?
A runner's speed is $v(t) = 4t + 3$ m/s, starting at position $0$ . How far has the runner gone after $4$ seconds?

Answers

$x^{3} + 2 x^{2} - 5 x + C$
$-\dfrac{5}{2x^2} - \dfrac{4}{3}x^{3/2} + C$
$e^{x} + 3 \sin x + C$
$x + x^{2} + \frac{x^{3}}{3} + C$
$\frac{x^{2}}{2} - \frac{2}{x} + C$
$\frac{x}{2} - \frac{\sin x}{2} + C$
$\tan^{-1} x + 2 \sin^{-1} x + C$
$\dfrac{2^x}{\ln 2} + \ln|x| + C$
$\tan^2 x + 1 = \sec^2 x$ , so the integral is $\tan x + C$ .
$y = x^3 - x^2 + C$ . At $(0, 1)$ : $C = 1$ . So $y = x^3 - x^2 + 1$ , and at $x = 2$ , $y = 8 - 4 + 1 = 5$ .
$s(t) = 2t^2 + 3t + C$ with $s(0) = 0$ , so $C = 0$ . Then $s(4) = 32 + 12 = 44$ metres.

7.3 Methods of Integration

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The basic table covers only a few functions. Most integrals need a method that turns them into basic ones. This chapter has four main methods. Here is a guide to choosing between them:

If the integrand looks like ...	Try ...	Section
a function of an inner function, with the derivative of the inner function nearby, such as $x\,(x^2+3)^4$	Substitution	7.3.1
powers or products of sine and cosine, such as $\sin^2 x$ or $\sin 3x \cos x$	Trigonometric identities	7.3.2
$\dfrac{1}{ax^2 + bx + c}$ or $\dfrac{1}{\sqrt{ax^2 + bx + c}}$	Standard forms (complete the square)	7.4
a fraction of two polynomials, such as $\dfrac{3x + 1}{(x-1)(x+2)}$	Partial fractions	7.5
a product of two different kinds of function, such as $x\,e^x$ or $\ln x$	Integration by parts	7.6

In simple words: an integral is a puzzle. Each method is a way to rewrite the puzzle as something you already know how to solve. When one method does not work, try another.

7.3.1 Integration by substitution

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Substitution undoes the chain rule. If an integrand contains an inner function $g(x)$ and, nearby, its derivative $g'(x)$ , give the inner function a new name $u$ . This turns a messy integral into a simple one.

The steps:

Choose $u = g(x)$ (usually the inside of a bracket, a root, a power, or the bottom of a fraction).
Differentiate: $du = g'(x)\,dx$ . Solve for what you need, such as $x\,dx = \frac{du}{2}$ .
Rewrite the whole integral using only $u$ and $du$ . No $x$ may be left.
Integrate with respect to $u$ .
Put $x$ back: replace $u$ by $g(x)$ .

Find $\int 2x\,(x^2 + 3)^4\,dx$ . Let $u = x^2 + 3$ . Then $du = 2x\,dx$ , which is exactly the rest of the integrand. So

$\int u^4\,du = \frac{u^5}{5} + C = \frac{(x^2 + 3)^5}{5} + C.$

Find $\int \cos(3x + 1)\,dx$ . Let $u = 3x + 1$ , so $du = 3\,dx$ and $dx = \frac{du}{3}$ . Then $\int \cos u \cdot \frac{du}{3} = \frac{\sin u}{3} + C = \frac{\sin(3x + 1)}{3} + C$ .

A shortcut for a linear inside. If

\int f(x)\,dx = F(x) + C

, then

\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C

. Divide by the number multiplying

x

.

Find $\int \dfrac{x}{x^2 + 4}\,dx$ . Let $u = x^2 + 4$ , so $x\,dx = \frac{du}{2}$ . Then $\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| + C$ . Since $x^2 + 4 > 0$ , the answer is $\frac{1}{2}\ln(x^2 + 4) + C$ .

This pattern is so common that it is worth remembering:

$\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C.$

Find $\int \tan x\,dx = \int \dfrac{\sin x}{\cos x}\,dx$ . Let $u = \cos x$ , so $du = -\sin x\,dx$ . Then $-\int \frac{du}{u} = -\ln|\cos x| + C$ . (This equals $\ln|\sec x| + C$ .)

Find $\int \sin x \cos^2 x\,dx$ . Let $u = \cos x$ , so $du = -\sin x\,dx$ . Then $-\int u^2\,du = -\frac{\cos^3 x}{3} + C$ .

Find $\int \sin^3 x\,dx$ . Split off one sine: $\sin^3 x = \sin x\,(1 - \cos^2 x)$ . With $u = \cos x$ , $-\int (1 - u^2)\,du = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C$ .

$\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx$ : let $u = \sqrt{x}$ , so $du = \frac{dx}{2\sqrt{x}}$ and the integral is $2\int e^u\,du = 2e^{\sqrt{x}} + C$ .
$\displaystyle\int \frac{dx}{x \ln x}$ for $x > 1$ : let $u = \ln x$ , so $du = \frac{dx}{x}$ and the integral is $\int \frac{du}{u} = \ln|\ln x| + C$ .

Common mistakes: (1) leaving some

x

in an integral that should now be in

u

; (2) forgetting the factor from

du

, such as the

\frac{1}{3}

in Example 2; (3) forgetting to replace

u

by its

x

-expression at the end.

All six answers above were checked by differentiating them. For instance, the derivative of $\frac{(x^2 + 3)^5}{5}$ is $2x(x^2 + 3)^4$ , as it should be.

EXERCISE 7.2

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Find $\int (3x + 2)^5 \, dx$ .
Find $\int \frac{x}{1 + x^2} \, dx$ .
Find $\int \cos x \sin^4 x \, dx$ .
Find $\int e^{2x + 3} \, dx$ .
Find $\int \sec^2 3x \, dx$ .
Find $\int \frac{(\ln x)^2}{x} \, dx$ for $x > 0$ .
Find $\int \frac{1}{2x + 5} \, dx$ .
Find $\int x\sqrt{x + 1} \, dx$ . (Let $u = x + 1$ and write $x = u - 1$ .)
Find $\int \cos^3 x \, dx$ . (Split off one cosine and use $\cos^2 x = 1 - \sin^2 x$ .)
Find $\int \frac{e^x}{1 + e^x} \, dx$ .

Answers

$\frac{\left(3 x + 2\right)^{6}}{18} + C$
$\dfrac{\ln(1 + x^2)}{2} + C$
$\frac{\sin^{5} x}{5} + C$
$\dfrac{e^{2x + 3}}{2} + C$
$\dfrac{\tan 3x}{3} + C$
$\dfrac{(\ln x)^3}{3} + C$
$\dfrac{\ln|2x + 5|}{2} + C$
$\dfrac{2}{5}(x+1)^{5/2} - \dfrac{2}{3}(x+1)^{3/2} + C$
$\sin x - \dfrac{\sin^3 x}{3} + C$
$\ln(1 + e^x) + C$

7.3.2 Integration using trigonometric identities

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Products and powers of sine and cosine have no direct rule. But trigonometric identities turn them into sums of simple sines and cosines, which you can integrate term by term.

The identities you need (all verified numerically):

$\sin^2 A = \dfrac{1 - \cos 2A}{2}$ and $\cos^2 A = \dfrac{1 + \cos 2A}{2}$ .
$\sin A \cos B = \dfrac{1}{2}\big[\sin(A + B) + \sin(A - B)\big]$ .
$\cos A \cos B = \dfrac{1}{2}\big[\cos(A - B) + \cos(A + B)\big]$ .
$\sin A \sin B = \dfrac{1}{2}\big[\cos(A - B) - \cos(A + B)\big]$ .

$\int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C.$

Find $\int \sin 5x \cos 2x\,dx$ . With $A = 5x$ and $B = 2x$ : $\sin 5x \cos 2x = \frac{1}{2}(\sin 7x + \sin 3x)$ . Then

$\int \sin 5x \cos 2x\,dx = -\frac{\cos 7x}{14} - \frac{\cos 3x}{6} + C.$

$\cos 3x \cos x = \frac{1}{2}(\cos 2x + \cos 4x)$ , so $\int \cos 3x \cos x\,dx = \frac{\sin 2x}{4} + \frac{\sin 4x}{8} + C$ .

Common mistake:

\sin^2 x

does not integrate to

\frac{\sin^3 x}{3}

. That would be the integral of

\sin^2 x \cos x

(by substitution). To integrate a plain

\sin^2 x

you need the identity.

EXERCISE 7.3

150 words · 1 min read

Find $\int \sin^2 3x \, dx$ .
Find $\int \cos^2 2x \, dx$ .
Find $\int \sin 4x \cos x \, dx$ .
Find $\int \sin x \sin 3x \, dx$ .
Find $\int \frac{1 - \cos x}{1 + \cos x} \, dx$ . (Show first that it equals $\tan^2\frac{x}{2}$ .)
Find $\int \cos 2x \cos 4x \, dx$ .

Answers

$\dfrac{x}{2} - \dfrac{\sin 6x}{12} + C$
$\frac{x}{2} + \frac{\sin\left(4 x\right)}{8} + C$
$\sin 4x \cos x = \frac{1}{2}(\sin 5x + \sin 3x)$ , so the integral is $\frac{-\cos\left(5 x\right)}{10} - \frac{\cos\left(3 x\right)}{6} + C$
$\sin x \sin 3x = \frac{1}{2}(\cos 2x - \cos 4x)$ , so the integral is $\frac{\sin\left(2 x\right)}{4} - \frac{\sin\left(4 x\right)}{8} + C$
$\dfrac{1 - \cos x}{1 + \cos x} = \tan^2\dfrac{x}{2} = \sec^2\dfrac{x}{2} - 1$ . So the integral is $2\tan\dfrac{x}{2} - x + C$
$\cos 2x \cos 4x = \frac{1}{2}(\cos 2x + \cos 6x)$ , so the integral is $\frac{\sin\left(2 x\right)}{4} + \frac{\sin\left(6 x\right)}{12} + C$

7.4 Integrals of Some Particular Functions

676 words · 3 min read

Six integrals come up so often that they are worth learning as formulas. Each one comes from a derivative you already know. In each row, $a$ is a positive constant.

Integrand	Integral (add $+C$ )
$\dfrac{1}{x^2 - a^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{x - a}{x + a}\right\|$
$\dfrac{1}{a^2 - x^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{a + x}{a - x}\right\|$
$\dfrac{1}{x^2 + a^2}$	$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$
$\dfrac{1}{\sqrt{x^2 - a^2}}$	$\ln\left\|x + \sqrt{x^2 - a^2}\right\|$
$\dfrac{1}{\sqrt{a^2 - x^2}}$	$\sin^{-1}\dfrac{x}{a}$
$\dfrac{1}{\sqrt{x^2 + a^2}}$	$\ln\left(x + \sqrt{x^2 + a^2}\right)$

Each formula was checked by differentiating it, using $a = 3$ , $a = 2$ or $a = 3$ in turn. The second row is valid for $x \ne \pm a$ , and the fourth for $x > a$ or $x < -a$ .

How to tell which formula: look at the sign.

x^2 - a^2

with a minus gives a logarithm (when the fraction has no root).

x^2 + a^2

with a plus gives an inverse tangent. A root with

a^2 - x^2

gives an inverse sine.

Real problems rarely arrive as $x^2 + a^2$ . They arrive as $x^2 + 4x + 13$ . The fix is to complete the square, which rewrites a quadratic as $(x + p)^2 \pm q^2$ , and then use the substitution $u = x + p$ .

Example 1. Find $\int \dfrac{dx}{x^2 + 4x + 13}$ . Complete the square: $x^2 + 4x + 13 = (x + 2)^2 + 9$ . With $u = x + 2$ and $a = 3$ :

$\int \frac{dx}{(x + 2)^2 + 3^2} = \frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C.$

Example 2. Find $\int \dfrac{dx}{x^2 - 6x + 5}$ . Here $x^2 - 6x + 5 = (x - 3)^2 - 2^2$ . With $u = x - 3$ and $a = 2$ (the "minus" formula):

$\frac{1}{4}\ln\left|\frac{x - 5}{x - 1}\right| + C.$

Example 3. Find $\int \dfrac{dx}{\sqrt{5 + 4x - x^2}}$ . Complete the square with a minus sign: $5 + 4x - x^2 = 9 - (x - 2)^2$ . This is $a^2 - u^2$ with $a = 3$ , so the answer is $\sin^{-1}\dfrac{x - 2}{3} + C$ .

Example 4. $\int \dfrac{dx}{\sqrt{x^2 + 2x + 10}}$ : since $x^2 + 2x + 10 = (x + 1)^2 + 9$ , the answer is $\ln\left(x + 1 + \sqrt{x^2 + 2x + 10}\right) + C$ .

When the numerator is $px + q$ , write it as (a multiple of the derivative of the denominator) plus a constant. The first part gives a logarithm or a root. The second part is one of the forms above.

Example 5. Find $\int \dfrac{2x + 5}{x^2 + 4x + 13}\,dx$ . The derivative of the denominator is $2x + 4$ , and $2x + 5 = (2x + 4) + 1$ . So

$\int \frac{2x + 4}{x^2 + 4x + 13}\,dx = \ln(x^2 + 4x + 13)$

$\int \frac{1}{x^2 + 4x + 13}\,dx = \frac{1}{3}\tan^{-1}\frac{x + 2}{3}$

Adding the two parts, the answer is $\ln(x^2 + 4x + 13) + \dfrac{1}{3}\tan^{-1}\dfrac{x + 2}{3} + C$ .

Example 6. Find $\int \dfrac{3x + 1}{\sqrt{x^2 + 2x + 5}}\,dx$ . The derivative of $x^2 + 2x + 5$ is $2x + 2$ , and $3x + 1 = \frac{3}{2}(2x + 2) - 2$ . The first part gives $3\sqrt{x^2 + 2x + 5}$ , the second gives a logarithm:

$3\sqrt{x^2 + 2x + 5} - 2\ln\left(x + 1 + \sqrt{x^2 + 2x + 5}\right) + C.$

Common mistake: using the

\tan^{-1}

formula without the

\frac{1}{a}

, or using

\ln

for

x^2 + a^2

. Differentiate your answer to be sure.

EXERCISE 7.4

154 words · 1 min read

Find $\int \frac{1}{x^2 + 16} \, dx$ .
Find $\int \frac{1}{25 - x^2} \, dx$ .
Find $\int \frac{1}{\sqrt{16 - 9x^2}} \, dx$ .
Find $\int \frac{1}{x^2 + 6x + 13} \, dx$ .
Find $\int \frac{1}{x^2 - 4x + 3} \, dx$ .
Find $\int \frac{1}{\sqrt{x^2 + 4x + 8}} \, dx$ .
Find $\int \frac{3x + 2}{x^2 + 2x + 5} \, dx$ .
Find $\int \frac{1}{\sqrt{1 - 4x^2}} \, dx$ .

Answers

$\frac{\tan^{-1}\left(\frac{x}{4}\right)}{4} + C$
$\dfrac{1}{10}\ln\left|\dfrac{5 + x}{5 - x}\right| + C$
$\dfrac{1}{3}\sin^{-1}\dfrac{3x}{4} + C$
$\dfrac{1}{2}\tan^{-1}\dfrac{x + 3}{2} + C$
$\dfrac{1}{2}\ln\left|\dfrac{x - 3}{x - 1}\right| + C$
$\ln\left(x + 2 + \sqrt{x^2 + 4x + 8}\right) + C$
$\dfrac{3}{2}\ln(x^2 + 2x + 5) - \dfrac{1}{2}\tan^{-1}\dfrac{x + 1}{2} + C$
$\dfrac{1}{2}\sin^{-1} 2x + C$

7.5 Integration by Partial Fractions

761 words · 3 min read

A rational function is a fraction of two polynomials, such as $\dfrac{5x + 1}{(x - 1)(x + 2)}$ . Integrating a single fraction like this is hard. But you can split it into simpler fractions that are each easy to integrate. This is the reverse of adding fractions with a common denominator. The simpler pieces are called partial fractions.

The method works when the fraction is proper: the top has a smaller degree than the bottom. If it is not proper, divide first (long division) and integrate the polynomial part and the proper fraction separately.

Factor in the bottom	Matching piece(s)
$(x - a)$	$\dfrac{A}{x - a}$
$(x - a)^2$ (repeated)	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$
$(x^2 + bx + c)$ that cannot be factorised	$\dfrac{Ax + B}{x^2 + bx + c}$

Write one piece for each factor, put unknown letters on top, multiply both sides by the common denominator, and find the letters. A fast way for a single factor $(x - a)$ : cover up that factor on the left, then put $x = a$ in what is left. This is the "cover-up" trick.

Find $\int \dfrac{5x + 1}{(x - 1)(x + 2)}\,dx$ . Write $\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ . Cover up $(x - 1)$ and put $x = 1$ : $A = \dfrac{5 + 1}{1 + 2} = 2$ . Cover up $(x + 2)$ and put $x = -2$ : $B = \dfrac{-10 + 1}{-2 - 1} = 3$ . So the integral of $\frac{2}{x - 1} + \frac{3}{x + 2}$ is

$2\ln|x - 1| + 3\ln|x + 2| + C.$

Find $\int \dfrac{3x + 1}{(x - 2)^2}\,dx$ . Write $3x + 1 = 3(x - 2) + 7$ , so $\dfrac{3x + 1}{(x - 2)^2} = \dfrac{3}{x - 2} + \dfrac{7}{(x - 2)^2}$ . Then, integrating each piece,

$3\ln|x - 2| - \frac{7}{x - 2} + C.$

Find $\int \dfrac{x^2 + 5}{(x - 1)(x - 2)(x + 1)}\,dx$ . Cover-up gives $A = \dfrac{1 + 5}{(1 - 2)(1 + 1)} = -3$ , $B = \dfrac{4 + 5}{(2 - 1)(2 + 1)} = 3$ , $C = \dfrac{1 + 5}{(-1 - 1)(-1 - 2)} = 1$ . So the integral is

$-3\ln|x - 1| + 3\ln|x - 2| + \ln|x + 1| + C.$

Find $\int \dfrac{x^2 + 2}{(x - 1)^2(x + 2)}\,dx$ . Write it as $\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2} + \dfrac{C}{x + 2}$ . Cover-up for the repeated factor at $x = 1$ gives $B = \dfrac{3}{3} = 1$ . Cover-up at $x = -2$ gives $C = \dfrac{6}{9} = \dfrac{2}{3}$ . Comparing the $x^2$ terms on both sides gives $A + C = 1$ , so $A = \dfrac{1}{3}$ . Then

$\int = \frac{1}{3}\ln|x - 1| - \frac{1}{x - 1} + \frac{2}{3}\ln|x + 2| + C.$

Find $\int \dfrac{dx}{(x - 1)(x^2 + 1)}$ . The quadratic $x^2 + 1$ cannot be factorised, so write $\dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 1}$ . Multiply out: $1 = A(x^2 + 1) + (Bx + C)(x - 1)$ . Putting $x = 1$ gives $A = \dfrac{1}{2}$ . Comparing $x^2$ : $A + B = 0$ , so $B = -\dfrac{1}{2}$ . Comparing constants: $A - C = 1$ , so $C = -\dfrac{1}{2}$ . The integral is

$\frac{1}{2}\ln|x - 1| - \frac{1}{4}\ln(x^2 + 1) - \frac{1}{2}\tan^{-1}x + C.$

Find $\int \dfrac{x^3 + 2x}{x^2 - 1}\,dx$ . The top has the bigger degree, so divide first: $x^3 + 2x = x(x^2 - 1) + 3x$ . So the fraction is $x + \dfrac{3x}{(x - 1)(x + 1)} = x + \dfrac{3/2}{x - 1} + \dfrac{3/2}{x + 1}$ , and the integral is

$\frac{x^2}{2} + \frac{3}{2}\ln|x^2 - 1| + C.$

Every splitting above was checked by computer at several values of $x$ (both sides equal), and every answer was checked by differentiating it.

Common mistakes: (1) trying to split an improper fraction before dividing; (2) writing

\frac{A}{x - a}

only, instead of both

\frac{A}{x - a}

and

\frac{B}{(x - a)^2}

, for a repeated factor; (3) forgetting the absolute value in the logarithms.

EXERCISE 7.5

181 words · 1 min read

Find $\int \frac{1}{(x + 1)(x + 3)} \, dx$ .
Find $\int \frac{3x + 1}{(x - 1)(x + 3)} \, dx$ .
Find $\int \frac{x}{(x - 1)^2} \, dx$ . (Write $x = (x - 1) + 1$ .)
Find $\int \frac{x^2 + 1}{x(x - 1)(x + 1)} \, dx$ .
Find $\int \frac{1}{x^2(x + 1)} \, dx$ .
Find $\int \frac{2x + 1}{(x + 1)(x^2 + 4)} \, dx$ .
Find $\int \frac{x^3}{x^2 + 1} \, dx$ . (Divide first.)
Find $\int \frac{1}{x^2 - 5x + 6} \, dx$ . (Factorise the bottom.)

Answers

$\dfrac{1}{2}\ln\left|\dfrac{x + 1}{x + 3}\right| + C$
$\ln|x - 1| + 2\ln|x + 3| + C$
$\ln|x - 1| - \dfrac{1}{x - 1} + C$
$\ln\left|\dfrac{x^2 - 1}{x}\right| + C$
$-\ln|x| - \dfrac{1}{x} + \ln|x + 1| + C$
$-\dfrac{1}{5}\ln|x + 1| + \dfrac{1}{10}\ln(x^2 + 4) + \dfrac{9}{10}\tan^{-1}\dfrac{x}{2} + C$
$\dfrac{x^2}{2} - \dfrac{1}{2}\ln(x^2 + 1) + C$
$\ln\left|\dfrac{x - 3}{x - 2}\right| + C$

7.6 Integration by Parts

559 words · 3 min read

Substitution undoes the chain rule. Integration by parts undoes the product rule. The product rule says $\dfrac{d}{dx}(uv) = u\,v' + u'\,v$ . Integrating both sides and rearranging gives

$\int u\,v'\,dx = u\,v - \int u'\,v\,dx.$

This turns an integral you cannot do into a different one that is (hopefully) easier. Use it for products of two different kinds of function, such as $x\,e^x$ or $x \sin x$ , and for functions like $\ln x$ and $\tan^{-1}x$ that you cannot integrate directly.

Choose $u$ as the factor that becomes simpler when you differentiate it. A memory aid, in order of priority, is ILATE:

Inverse trigonometric ( $\tan^{-1}x$ , $\sin^{-1}x$ )
Logarithmic ( $\ln x$ )
Algebraic ( $x$ , $x^2$ )
Trigonometric ( $\sin x$ , $\cos x$ )
Exponential ( $e^x$ )

The factor that comes first in the list is $u$ . The rest, with $dx$ , is $v'\,dx$ .

Find $\int x\,e^x\,dx$ . Take $u = x$ (algebraic comes before exponential) and $v' = e^x$ , so $u' = 1$ and $v = e^x$ . Then

$\int x\,e^x\,dx = x\,e^x - \int e^x\,dx = (x - 1)e^x + C.$

$\int x\sin x\,dx$ : $u = x$ , $v' = \sin x$ , $v = -\cos x$ . So $-x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C$ .

$\int \ln x\,dx$ looks like it has no product. Write it as $1 \cdot \ln x$ . Take $u = \ln x$ and $v' = 1$ , so $u' = \frac{1}{x}$ and $v = x$ . Then $x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C$ .

$\int x^2 e^x\,dx$ : first, $u = x^2$ gives $x^2 e^x - \int 2x\,e^x\,dx$ . The remaining integral is Example 1 (times $2$ ). So the answer is $x^2 e^x - 2(x - 1)e^x + C = e^x(x^2 - 2x + 2) + C$ .

Find $I = \int e^x \sin x\,dx$ . Take $u = \sin x$ , $v' = e^x$ : $I = e^x\sin x - \int e^x \cos x\,dx$ . Apply parts again to the new integral, with $u = \cos x$ : $\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x \cos x + I$ . Put this back:

$I = e^x \sin x - e^x \cos x - I \quad \Longrightarrow \quad 2I = e^x(\sin x - \cos x).$

So $\int e^x\sin x\,dx = \dfrac{e^x(\sin x - \cos x)}{2} + C$ . The trick is that the original integral appears again, and you solve for it like an equation.

$\int \tan^{-1}x\,dx$ : write it as $1 \cdot \tan^{-1}x$ , take $u = \tan^{-1}x$ , $v' = 1$ . Then $x\tan^{-1}x - \int \dfrac{x}{1 + x^2}\,dx = x\tan^{-1}x - \dfrac{1}{2}\ln(1 + x^2) + C$ .

$\int x\ln x\,dx$ : $u = \ln x$ (L before A), $v' = x$ , so $v = \frac{x^2}{2}$ . Then $\frac{x^2}{2}\ln x - \int \frac{x}{2}\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$ .

Common mistake: choosing

u

the wrong way round. For

\int x\,e^x\,dx

, taking

u = e^x

and

v' = x

makes the new integral

\int \frac{x^2}{2}e^x\,dx

, which is harder, not easier. If the new integral looks worse, swap the roles.

EXERCISE 7.6

128 words · 1 min read

Find $\int x\cos x \, dx$ .
Find $\int x\,e^{2x} \, dx$ .
Find $\int x^2\ln x \, dx$ .
Find $\int x\sin 2x \, dx$ .
Find $\int \sin^{-1}x \, dx$ .
Find $\int x^2\cos x \, dx$ .
Find $\int e^x\cos x \, dx$ . (The integral comes back.)
Find $\int (\ln x)^2 \, dx$ . (Write it as $1 \cdot (\ln x)^2$ and use parts twice.)

Answers

$x \sin x + \cos x + C$
$\dfrac{e^{2x}(2x - 1)}{4} + C$
$\dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$
$-\dfrac{x\cos 2x}{2} + \dfrac{\sin 2x}{4} + C$
$x \sin^{-1} x + \sqrt{1 - x^{2}} + C$
$x^{2} \sin x + 2 x \cos x - 2 \sin x + C$
$\dfrac{e^x(\cos x + \sin x)}{2} + C$
$x(\ln x)^2 - 2x\ln x + 2x + C$

7.6.2 Integrals of some more types

385 words · 2 min read

Sometimes an integrand has the form $e^x\big[f(x) + f'(x)\big]$ : a function plus its own derivative, multiplied by $e^x$ . The product rule shows what to expect: $\dfrac{d}{dx}\big(e^x f(x)\big) = e^x f(x) + e^x f'(x)$ . So

$\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C.$

$\int e^x(\sin x + \cos x)\,dx = e^x \sin x + C$ , because $\cos x$ is the derivative of $\sin x$ (here $f = \sin x$ ).
$\int e^x\left(\dfrac{1}{x} - \dfrac{1}{x^2}\right)dx = \dfrac{e^x}{x} + C$ , because $-\frac{1}{x^2}$ is the derivative of $\frac{1}{x}$ (here $f = \frac{1}{x}$ ).
$\int e^x(1 + x)\,dx = x\,e^x + C$ (here $f = x$ , $f' = 1$ ).

How to spot it: the bracket has two parts, and one is the derivative of the other. If you can match them, the answer is

e^x

times the original one.

Three integrals of square roots are worth knowing. They are found by integration by parts, taking $1 \cdot \sqrt{\cdots}$ as the product. In each, $a > 0$ .

Integrand	Integral (add $+C$ )
$\sqrt{a^2 - x^2}$	$\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}$
$\sqrt{x^2 + a^2}$	$\dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\sqrt{x^2 - a^2}$	$\dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\ln\left\|x + \sqrt{x^2 - a^2}\right\|$

Each formula was checked by differentiating it for a specific value of $a$ (that is $3$ , $2$ , $2$ in turn).

Example. Find $\int \sqrt{x^2 + 2x + 5}\,dx$ . Complete the square: $x^2 + 2x + 5 = (x + 1)^2 + 2^2$ . With $u = x + 1$ and $a = 2$ , use the second formula:

$\frac{x + 1}{2}\sqrt{x^2 + 2x + 5} + 2\ln\left(x + 1 + \sqrt{x^2 + 2x + 5}\right) + C.$

Geometric meaning. $\int_{-a}^{a}\sqrt{a^2 - x^2}\,dx$ is the area of a half-disc of radius $a$ . You will see this in Section 7.8, and the formula above gives $\frac{\pi a^2}{2}$ , as expected.

EXERCISE 7.7

124 words · 1 min read

Find $\int e^x(\cos x - \sin x) \, dx$ .
Find $\int e^x(x^2 + 2x) \, dx$ .
Find $\int e^x\left(\frac{1}{x} + \ln x\right) \, dx$ .
Find $\int e^x\,\frac{x - 1}{x^2} \, dx$ .
Find $\int \sqrt{16 - x^2} \, dx$ .
Find $\int \sqrt{x^2 + 9} \, dx$ .
Find $\int \sqrt{x^2 - 9} \, dx$ .

Answers

$e^{x} \cos x + C$
$x^{2} e^{x} + C$
$e^{x} \ln x + C$
$\dfrac{e^x}{x} + C$
$\dfrac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\dfrac{x}{4} + C$
$\dfrac{x}{2}\sqrt{x^2 + 9} + \dfrac{9}{2}\ln\left(x + \sqrt{x^2 + 9}\right) + C$
$\dfrac{x}{2}\sqrt{x^2 - 9} - \dfrac{9}{2}\ln\left|x + \sqrt{x^2 - 9}\right| + C$

7.7 Definite Integral

501 words · 2 min read

So far, an integral gave a function, with a $+C$ . A definite integral gives a number. It is written $\int_a^b f(x)\,dx$ and means the total amount of $f$ collected as $x$ moves from $a$ to $b$ . The numbers $a$ and $b$ are the limits of integration.

When $f(x) \ge 0$ , this total is the area between the graph of $f$ , the $x$ -axis, and the lines $x = a$ and $x = b$ . How can we calculate such an area when the top edge is curved? We trap it between rectangles.

Let us find the area under $y = x^2$ from $x = 0$ to $x = 2$ . Cut $[0, 2]$ into $n$ equal strips of width $h = \frac{2}{n}$ . On each strip draw a rectangle whose height is the value of $x^2$ at the left end (a left sum), or at the right end (a right sum). The picture shows $8$ strips with left-end heights.

Graph of the curve y equals x squared from 0 to 2, with eight narrow rectangles under it. The top of each rectangle touches the curve at its left corner, so all eight rectangles lie below the curve and leave small gaps.

Eight left-end rectangles under y = x². Their total area is smaller than the true area, because each one sits below the curve.

With more strips the rectangles fit the curve better. Here are the totals (computed by computer):

Strips $n$	Left sum	Right sum	Middle sum
$4$	$1.75$	$3.75$	$2.625$
$10$	$2.28$	$3.08$	$2.66$
$100$	$2.6268$	$2.7068$	$2.6666$
$1000$	$2.6627$	$2.6707$	$2.6667$

The left sums rise and the right sums fall, and both squeeze towards the same number, $\frac{8}{3} \approx 2.6667$ . That number is the exact area. In general, the definite integral is defined as this limit:

$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,h, \qquad h = \frac{b - a}{n}.$

The sum sign $\sum$ means "add up", and the integral sign $\int$ is a stretched "S" for "sum". So $\int_0^2 x^2\,dx = \frac{8}{3}$ .

If $f$ is negative, the rectangles have negative height, so that part counts as negative area. The picture shows $y = \sin x$ from $0$ to $2\pi$ . The hump above the axis has area $2$ . The hump below has "area" $-2$ . They cancel, so $\int_0^{2\pi}\sin x\,dx = 0$ .

Graph of the sine wave from 0 to 2 pi. The first hump, above the axis, is shaded teal. The second hump, below the axis, is shaded orange. The two humps are equal in size.

The teal hump counts as +2 and the orange hump as −2, so the integral over the full period is 0.

Common mistake: treating a definite integral as always "the area". It equals the area only when

f \ge 0

. To find the total (unsigned) area of a graph that crosses the axis, split the interval at each crossing and add the absolute values.

7.8 Fundamental Theorem of Calculus

51 words · 1 min read

Adding up thousands of strips is slow. Fortunately two theorems turn the whole job into one subtraction. They are so important that they are called fundamental. The first says that the accumulated area is an antiderivative. The second uses that fact to calculate any definite integral.

7.8.1 Area function

238 words · 1 min read

Fix a starting point $a$ and let the right-hand end move. The area collected from $a$ up to a variable point $x$ is a function of $x$ . It is called the area function:

$A(x) = \int_a^x f(t)\,dt.$

(We use the letter $t$ inside, so that $x$ is free to be the upper limit.) As $x$ moves to the right, $A(x)$ grows by the strip of area that is added.

Example. Let $f(t) = 2t + 1$ and $a = 0$ . The region under this line from $0$ to $x$ is a trapezoid, with area $\frac{1}{2}\big(1 + (2x + 1)\big)x = x^2 + x$ . So $A(x) = x^2 + x$ . The computer confirmed this by integrating numerically:

$x$	$1$	$2$	$3$	$4$
$A(x) = x^2 + x$	$2$	$6$	$12$	$20$

Graph of the straight line y equals 2 t plus 1 from t equals 0 to 2. The region under the line, from 0 to 2, is shaded. The line starts at height 1 and ends at height 5.

A(2) is the shaded area under y = 2t + 1 from 0 to 2. It equals 2² + 2 = 6.

Now look at the derivative of the area function: $A'(x) = 2x + 1$ . That is exactly $f(x)$ , the height of the line at $x$ . This is not an accident. It is the First Fundamental Theorem.

7.8.2 First fundamental theorem of integral calculus

248 words · 1 min read

Theorem. If $f$ is continuous on $[a, b]$ , then the area function $A(x) = \int_a^x f(t)\,dt$ is differentiable, and

$A'(x) = f(x).$

In words: the rate at which the area grows is the height of the graph. If the graph is high at $x$ , the area grows fast there. If the graph is at height zero, the area does not grow.

Why it is true. Move the right end from $x$ to $x + h$ . The area added is a thin strip of width $h$ and height about $f(x)$ , so $A(x + h) - A(x) \approx f(x)\,h$ . Divide by $h$ and let $h \to 0$ : $\dfrac{A(x + h) - A(x)}{h} \to f(x)$ .

Consequence. Every continuous function $f$ has an antiderivative, namely its area function. This is the link between the two meanings of "integral".

$\dfrac{d}{dx}\displaystyle\int_1^x t^2\,dt = x^2$ . You do not need to integrate first. (If you do: the integral is $\frac{x^3}{3} - \frac{1}{3}$ , whose derivative is $x^2$ .)
If the upper limit is a function of $x$ , use the chain rule: $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt = \cos(x^2)\cdot 2x$ . As a check, $\int_0^{x^2}\cos t\,dt = \sin(x^2)$ , whose derivative is $2x\cos(x^2)$ . The computer confirmed $\int_0^4 \cos t\,dt = \sin 4$ .

Common mistake: forgetting the chain-rule factor when the limit is not just

x

. Differentiating

\int_0^{x^2} \cos t\,dt

gives

2x\cos(x^2)

, not

\cos(x^2)

.

7.8.3 Second fundamental theorem of integral calculus

401 words · 2 min read

Theorem. Let $f$ be continuous on $[a, b]$ , and let $F$ be any antiderivative of $f$ . Then

$\int_a^b f(x)\,dx = F(b) - F(a).$

This turns the limit of sums into a subtraction. We write the subtraction as $\Big[F(x)\Big]_a^b$ , which means "put in $b$ , put in $a$ , subtract".

Why it is true. The area function $A(x) = \int_a^x f(t)\,dt$ is one antiderivative. Any other antiderivative $F$ differs from it by a constant: $F(x) = A(x) + C$ . Then $F(b) - F(a) = \big(A(b) + C\big) - \big(A(a) + C\big) = A(b) - A(a) = A(b) - 0$ . The constant $C$ cancels. That is why you never need $+C$ in a definite integral.

Find an antiderivative $F$ (any one, take $C = 0$ ).
Calculate $F(b) - F(a)$ .
Write the answer as a number.

$\displaystyle\int_1^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_1^3 = 9 - \dfrac{1}{3} = \dfrac{26}{3}$ .
$\displaystyle\int_0^{\pi}\sin x\,dx = \Big[-\cos x\Big]_0^{\pi} = 1 - (-1) = 2$ . This is the area of one hump of the sine wave.
$\displaystyle\int_1^{e}\dfrac{dx}{x} = \Big[\ln x\Big]_1^{e} = 1 - 0 = 1$ .
$\displaystyle\int_0^1\dfrac{dx}{1 + x^2} = \Big[\tan^{-1}x\Big]_0^1 = \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}$ .
$\displaystyle\int_1^4\sqrt{x}\,dx = \left[\dfrac{2}{3}x^{3/2}\right]_1^4 = \dfrac{16}{3} - \dfrac{2}{3} = \dfrac{14}{3}$ .
A half-disc. By the formula from Section 7.6.2, $\displaystyle\int_{-3}^{3}\sqrt{9 - x^2}\,dx = \left[\dfrac{x}{2}\sqrt{9 - x^2} + \dfrac{9}{2}\sin^{-1}\dfrac{x}{3}\right]_{-3}^{3} = \dfrac{9\pi}{4} - \left(-\dfrac{9\pi}{4}\right) = \dfrac{9\pi}{2}$ . This is the area of a half-disc of radius $3$ .

Each of these six values was also checked by integrating numerically (adding thousands of thin strips), and they agree: $8.6667$ , $2$ , $1$ , $0.7854$ , $14.1372$ , $4.6667$ .

The theorem needs a continuous function. Try

\int_{-1}^{1}\frac{dx}{x^2}

with

F(x) = -\frac{1}{x}

. Then

F(1) - F(-1) = -1 - 1 = -2

. But

\frac{1}{x^2}

is always positive (for example it equals

4

at

x = 0.5

), so its integral cannot be negative. The mistake:

\frac{1}{x^2}

is not defined at

x = 0

, so it is not continuous on

[-1, 1]

.

In simple words: to find the total change, you do not have to add up every tiny piece. You only need the antiderivative at the two end points.

EXERCISE 7.8

134 words · 1 min read

Evaluate $\int_{0}^{2} 3 x^{2} + 1 \, dx$ .
Evaluate $\int_{1}^{4} x + \dfrac{1}{\sqrt{x}} \, dx$ .
Evaluate $\int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx$ .
Evaluate $\int_{0}^{1} e^{2 x} \, dx$ .
Evaluate $\int_{1}^{2} \frac{1}{x^{2}} \, dx$ .
Evaluate $\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} \, dx$ .
Evaluate $\int_{2}^{3} \frac{x}{x^{2} - 1} \, dx$ . (Use $u = x^2 - 1$ or partial fractions.)
Evaluate $\int_{0}^{3} \sqrt{9 - x^2} \, dx$ . (A quarter-disc of radius $3$ . Use the formula from 7.6.2.)
Evaluate $\int_{0}^{1} x\,e^x \, dx$ . (Integrate by parts.)
Evaluate $\int_{1}^{e} \ln x \, dx$ . (Integrate by parts.)

Answers

$10 \approx 10$
$\dfrac{19}{2} \approx 9.5$
$1 \approx 1$
$\dfrac{e^2 - 1}{2} \approx 3.1945$
$\dfrac{1}{2} \approx 0.5$
$\dfrac{\pi}{6} \approx 0.5236$
$\dfrac{1}{2}\ln\dfrac{8}{3} \approx 0.4904$
$\dfrac{9\pi}{4} \approx 7.0686$
$1 \approx 1$
$1 \approx 1$

7.9 Evaluation of Definite Integrals by Substitution

348 words · 2 min read

Substitution works for definite integrals too, with one extra step that saves work: change the limits as you change the variable. Then you never have to go back to $x$ .

The steps:

Choose $u = g(x)$ and find $du$ , as before.
Change the limits: when $x = a$ , $u = g(a)$ . When $x = b$ , $u = g(b)$ .
Rewrite the integral in terms of $u$ , with the new limits.
Integrate and subtract: $F(\text{new upper}) - F(\text{new lower})$ .

Find $\displaystyle\int_0^1 2x\,(x^2 + 1)^3\,dx$ . Let $u = x^2 + 1$ , so $du = 2x\,dx$ . When $x = 0$ , $u = 1$ . When $x = 1$ , $u = 2$ . So

$\int_1^2 u^3\,du = \left[\frac{u^4}{4}\right]_1^2 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}.$

Find $\displaystyle\int_0^{\pi/2}\sin x\cos^3 x\,dx$ . Let $u = \cos x$ , so $du = -\sin x\,dx$ . At $x = 0$ , $u = 1$ . At $x = \frac{\pi}{2}$ , $u = 0$ . The integral becomes $-\int_1^0 u^3\,du = \int_0^1 u^3\,du = \dfrac{1}{4}$ . (Swapping the limits changes the sign, so the minus sign disappears.)

Find $\displaystyle\int_0^1\dfrac{x}{1 + x^2}\,dx$ . Let $u = 1 + x^2$ , so $x\,dx = \frac{du}{2}$ . Limits: $u$ goes from $1$ to $2$ . So $\frac{1}{2}\int_1^2\frac{du}{u} = \frac{1}{2}\ln 2$ .

Find $\displaystyle\int_0^4\dfrac{dx}{1 + \sqrt{x}}$ . Let $u = \sqrt{x}$ , so $x = u^2$ and $dx = 2u\,du$ . Limits: $u$ goes from $0$ to $2$ . Then $\int_0^2\dfrac{2u}{1 + u}\,du = 2\int_0^2\left(1 - \dfrac{1}{1 + u}\right)du = 2\Big[u - \ln(1 + u)\Big]_0^2 = 4 - 2\ln 3$ .

All four answers were checked numerically, both in the original form and in the new form with the new limits.

Common mistake: changing the variable but keeping the old limits. If you write

\int_0^1 u^3\,du

for Example 1, you get

\frac{1}{4}

, which is wrong. The limits must belong to the new variable.

EXERCISE 7.9

139 words · 1 min read

Evaluate $\int_{0}^{1} x(x^2 + 4)^3 \, dx$ . ( $u = x^2 + 4$ )
Evaluate $\int_{0}^{\frac{\pi}{2}} \cos x \sin^2 x \, dx$ . ( $u = \sin x$ )
Evaluate $\int_{0}^{1} \dfrac{e^x}{1 + e^x} \, dx$ . ( $u = 1 + e^x$ )
Evaluate $\int_{0}^{3} x\sqrt{x + 1} \, dx$ . ( $u = x + 1$ )
Evaluate $\int_{e}^{e^2} \dfrac{1}{x\ln x} \, dx$ . ( $u = \ln x$ )
Evaluate $\int_{0}^{\frac{\pi}{2}} \dfrac{\sin 2x}{1 + \cos^2 x} \, dx$ . ( $u = 1 + \cos^2 x$ ; note $\sin 2x = 2\sin x\cos x$ )
Evaluate $\int_{0}^{2} x^2\sqrt{1 + x^3} \, dx$ . ( $u = 1 + x^3$ )
Evaluate $\int_{0}^{1} \dfrac{1}{1 + e^{-x}} \, dx$ . (Multiply top and bottom by $e^x$ .)

Answers

$\dfrac{369}{8} \approx 46.125$
$\dfrac{1}{3} \approx 0.3333$
$\ln\dfrac{1 + e}{2} \approx 0.6201$
$\dfrac{116}{15} \approx 7.7333$
$\ln 2 \approx 0.6931$
$\ln 2 \approx 0.6931$
$\dfrac{52}{9} \approx 5.7778$
$\ln\dfrac{1 + e}{2} \approx 0.6201$

7.10 Some Properties of Definite Integrals

499 words · 2 min read

These rules make many definite integrals easy, sometimes without finding an antiderivative at all. Each rule was checked on an example by computer.

The letter does not matter. $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ . The variable inside is a dummy.
Reversing the limits flips the sign. $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ . Example: $\int_0^3 x^2\,dx = 9$ , and $\int_3^0 x^2\,dx = -9$ . Also $\int_a^a f(x)\,dx = 0$ .
Splitting. For any $c$ , $\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ . Example: $\int_0^3 x^2\,dx = \frac{1}{3} + \frac{26}{3} = 9$ (split at $c = 1$ ). The point $c$ may even lie outside $[a, b]$ . This is how you integrate a function defined in pieces.
Linearity. $\int_a^b\big(k\,f + m\,g\big)\,dx = k\int_a^b f\,dx + m\int_a^b g\,dx$ .

Rule A. $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ . This reflects the graph left to right across the middle of $[a, b]$ . The area does not change. Example: on $[1, 3]$ , $f(x) = x$ becomes $f(4 - x) = 4 - x$ . Both integrals equal $4$ .

Rule B. The special case $a = 0$ : $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ .

Find $I = \displaystyle\int_0^{\pi/2}\dfrac{\sin x}{\sin x + \cos x}\,dx$ . There is no easy antiderivative. Use Rule B with $a = \frac{\pi}{2}$ : replacing $x$ by $\frac{\pi}{2} - x$ swaps sine and cosine, so $I = \displaystyle\int_0^{\pi/2}\dfrac{\cos x}{\sin x + \cos x}\,dx$ . Add the two forms:

$2I = \int_0^{\pi/2}\frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}.$

So $I = \dfrac{\pi}{4} \approx 0.7854$ . The computer confirms both $\int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx$ and the cosine version give $0.7854$ .

Rule C. For $\int_0^{2a}$ : $\displaystyle\int_0^{2a} f(x)\,dx = 2\int_0^{a} f(x)\,dx$ if $f(2a - x) = f(x)$ , and $= 0$ if $f(2a - x) = -f(x)$ . Example with $a = \pi$ : $\sin^2(2\pi - x) = \sin^2 x$ , so $\int_0^{2\pi}\sin^2 x\,dx = 2\int_0^{\pi}\sin^2 x\,dx = 2\cdot\frac{\pi}{2} = 3.1416$ . And $\sin(2\pi - x) = -\sin x$ , so $\int_0^{2\pi}\sin x\,dx = 0$ .

Rule D (even and odd). An even function has $f(-x) = f(x)$ (its graph is symmetric about the vertical axis). An odd function has $f(-x) = -f(x)$ . Then

even: $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^{a} f(x)\,dx$ . Example: $\int_{-2}^{2}x^2\,dx = \frac{16}{3} = 2 \cdot \frac{8}{3}$ .
odd: $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$ . The left half cancels the right half. Example: $\int_{-2}^{2}x^3\,dx = 0$ .

How to use the symmetry rules: before you integrate over

[-a, a]

, ask "is it even or odd?" An odd function gives

0

at once, with no calculation.

Common mistake: using the even/odd rule when the limits are not symmetric (

-a

to

a

), or when the function is neither even nor odd, such as

x^2 + x

.

EXERCISE 7.10

122 words · 1 min read

Evaluate $\int_{-1}^{1} x^5\cos x \, dx$ . (Is it even or odd?)
Evaluate $\int_{0}^{\frac{\pi}{2}} \dfrac{\cos x}{\sin x + \cos x} \, dx$ . (Use Rule B and add.)
Evaluate $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx$ . (Even function.)
Evaluate $\int_{0}^{1} x(1 - x)^5 \, dx$ . (Use Rule B: replace $x$ by $1 - x$ .)
Evaluate $\int_{1}^{4} \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{5 - x}} \, dx$ . (Use Rule A, then add the two forms.)
Evaluate $\int_{-2}^{2} |x| \, dx$ . (Even function. Or split at $0$ .)
Evaluate $\int_{0}^{2\pi} \cos^3 x \, dx$ . (Compare $\cos^3(2\pi - x)$ with $\cos^3 x$ .)
Evaluate $\int_{-\pi}^{\pi} x^2 \sin x \, dx$ . (Odd function.)

Answers

$0 \approx 0$
$\dfrac{\pi}{4} \approx 0.7854$
$\dfrac{\pi}{2} \approx 1.5708$
$\dfrac{1}{42} \approx 0.0238$
$\dfrac{3}{2} \approx 1.5$
$4 \approx 4$
$0 \approx 0$
$0 \approx 0$

Miscellaneous Examples

316 words · 1 min read

Find $\displaystyle\int\dfrac{dx}{e^x + e^{-x}}$ . Multiply top and bottom by $e^x$ to get $\dfrac{e^x}{e^{2x} + 1}$ . Let $u = e^x$ , so $du = e^x\,dx$ . The integral becomes $\displaystyle\int\dfrac{du}{u^2 + 1} = \tan^{-1}u + C = \tan^{-1}(e^x) + C$ .

Find $\displaystyle\int\dfrac{\sin x}{1 + \cos^2 x}\,dx$ . Let $u = \cos x$ , so $du = -\sin x\,dx$ . Then $-\displaystyle\int\dfrac{du}{1 + u^2} = -\tan^{-1}(\cos x) + C$ .

Find $\displaystyle\int\dfrac{dx}{x(x^2 + 1)}$ . Split: $\dfrac{1}{x(x^2 + 1)} = \dfrac{1}{x} - \dfrac{x}{x^2 + 1}$ . So the integral is $\ln|x| - \dfrac{1}{2}\ln(x^2 + 1) + C$ .

Find $\displaystyle\int_0^{\pi} x\sin x\,dx$ . By parts, an antiderivative is $-x\cos x + \sin x$ . Then $\Big[-x\cos x + \sin x\Big]_0^{\pi} = (\pi + 0) - (0 + 0) = \pi$ .

A tank takes in water at the rate $q(t) = 12t - 3t^2$ litres per minute for $0 \le t \le 4$ . How much water enters in these $4$ minutes? The total is $\displaystyle\int_0^4(12t - 3t^2)\,dt = \Big[6t^2 - t^3\Big]_0^4 = 96 - 64 = 32$ litres.

Find $\displaystyle\int_{-1}^{1}(x^4 + x^3)\,dx$ . The term $x^3$ is odd, so it contributes $0$ . The term $x^4$ is even, so it gives $2\displaystyle\int_0^1 x^4\,dx = 2 \cdot \dfrac{1}{5} = \dfrac{2}{5}$ .

Find $\displaystyle\int_0^1\tan^{-1}x\,dx$ . From Example 6 of Section 7.6, an antiderivative is $x\tan^{-1}x - \dfrac{1}{2}\ln(1 + x^2)$ . Then the value is $\left(\dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\right) - 0 \approx 0.4388$ .

Miscellaneous Exercise on Chapter 7

157 words · 1 min read

Find $\int \frac{x}{x^4 + 1} \, dx$ . ( $u = x^2$ )
Find $\int e^x\,\frac{1 + \sin x}{1 + \cos x} \, dx$ . (Show that the fraction equals $\tan\frac{x}{2} + \frac{1}{2}\sec^2\frac{x}{2}$ , then use $e^x[f + f']$ .)
Find $\int x^2 e^{x^3} \, dx$ . ( $u = x^3$ )
Find $\int \frac{1}{1 + \cos x} \, dx$ . ( $1 + \cos x = 2\cos^2\frac{x}{2}$ )
Find $\int x\sin x\cos x \, dx$ . (Use $\sin x\cos x = \frac{1}{2}\sin 2x$ , then parts.)
Evaluate $\int_{0}^{\frac{\pi}{4}} \tan x \, dx$ .
Evaluate $\int_{0}^{1} x\sqrt{1 - x^2} \, dx$ . ( $u = 1 - x^2$ )
Evaluate $\int_{0}^{2} |x - 1| \, dx$ . (Split at $x = 1$ : the graph is a V.)

Answers

$\dfrac{1}{2}\tan^{-1}(x^2) + C$
$e^x\tan\dfrac{x}{2} + C$
$\dfrac{e^{x^3}}{3} + C$
$\tan\dfrac{x}{2} + C$
$-\dfrac{x\cos 2x}{4} + \dfrac{\sin 2x}{8} + C$
$\dfrac{1}{2}\ln 2 \approx 0.3466$
$\dfrac{1}{3} \approx 0.3333$
$1 \approx 1$

Summary

214 words · 1 min read

Antiderivative. $F$ is an antiderivative of $f$ if $F' = f$ . The indefinite integral is $\int f(x)\,dx = F(x) + C$ . Always check by differentiating.
Linearity. $\int(af + bg)\,dx = a\int f\,dx + b\int g\,dx$ . The integral of a product is not the product of integrals.
Substitution. Let $u = g(x)$ , $du = g'(x)\,dx$ . For a definite integral, change the limits too. A shortcut: $\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C$ .
Trigonometric identities turn products and squares of sine and cosine into sums.
Partial fractions split a proper rational function into simple fractions. Divide first if it is not proper.
By parts: $\int u\,v'\,dx = uv - \int u'v\,dx$ . Choose $u$ by ILATE. Also $\int e^x[f + f']\,dx = e^x f + C$ .
Definite integral. $\int_a^b f(x)\,dx$ is the limit of Riemann sums. It is a signed area.
First Fundamental Theorem. If $A(x) = \int_a^x f(t)\,dt$ , then $A'(x) = f(x)$ .
Second Fundamental Theorem. $\int_a^b f(x)\,dx = F(b) - F(a)$ for any antiderivative $F$ of a continuous $f$ .
Properties. Reverse limits flips the sign. Split at any point. $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ . For even $f$ : $\int_{-a}^a f = 2\int_0^a f$ . For odd $f$ : $\int_{-a}^a f = 0$ .

Summary: partial fractions

145 words · 1 min read

Fraction	Split into
$\dfrac{px + q}{(x - a)(x - b)}$	$\dfrac{A}{x - a} + \dfrac{B}{x - b}$
$\dfrac{px + q}{(x - a)^2}$	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$
$\dfrac{px^2 + qx + r}{(x - a)(x - b)(x - c)}$	$\dfrac{A}{x - a} + \dfrac{B}{x - b} + \dfrac{C}{x - c}$
$\dfrac{px^2 + qx + r}{(x - a)^2(x - b)}$	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2} + \dfrac{C}{x - b}$
$\dfrac{px^2 + qx + r}{(x - a)(x^2 + bx + c)}$	$\dfrac{A}{x - a} + \dfrac{Bx + C}{x^2 + bx + c}$

Remember: the top must have a smaller degree than the bottom. If not, divide first.

Summary: integration by substitution

70 words · 1 min read

If the integrand contains ...	try $u = $
$f(ax + b)$	$ax + b$
$(\text{expression})^n$ and the derivative of the expression	the expression
$\dfrac{f'(x)}{f(x)}$	$f(x)$ (answer: $\ln\|f(x)\|$ )
$e^{g(x)}\,g'(x)$	$g(x)$
$\sqrt{x}$ or $\sqrt[n]{ax + b}$	the root

Rewrite with only $u$ and $du$ , integrate, then replace $u$ . For definite integrals, change the limits instead.

Summary: integrals of some special functions

105 words · 1 min read

Integrand	Integral (add $+C$ )
$\dfrac{1}{x^2 - a^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{x - a}{x + a}\right\|$
$\dfrac{1}{a^2 - x^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{a + x}{a - x}\right\|$
$\dfrac{1}{x^2 + a^2}$	$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$
$\dfrac{1}{\sqrt{x^2 - a^2}}$	$\ln\left\|x + \sqrt{x^2 - a^2}\right\|$
$\dfrac{1}{\sqrt{a^2 - x^2}}$	$\sin^{-1}\dfrac{x}{a}$
$\dfrac{1}{\sqrt{x^2 + a^2}}$	$\ln\left(x + \sqrt{x^2 + a^2}\right)$

For a general quadratic, complete the square first.

Summary: integration by parts

51 words · 1 min read

$\int u\,v'\,dx = u\,v - \int u'\,v\,dx$

Choose $u$ in the order Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.
For $\ln x$ or $\tan^{-1}x$ alone, take $v' = 1$ .
If the original integral comes back (as in $\int e^x\sin x\,dx$ ), solve for it like an equation.

Summary: some special types of integrals

124 words · 1 min read

$\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C$ .
$\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ .
$\int\sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left(x + \sqrt{x^2 + a^2}\right) + C$ .
$\int\sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C$ .
For a linear top $px + q$ over a quadratic or a square root of it: write $px + q$ as a multiple of the derivative of the denominator, plus a constant.

Summary: first fundamental theorem of integral calculus

54 words · 1 min read

If $f$ is continuous and $A(x) = \int_a^x f(t)\,dt$ , then $A'(x) = f(x)$ . The rate at which the area grows equals the height of the graph. With a function of $x$ as the upper limit, use the chain rule: $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$ .

Summary: second fundamental theorem of integral calculus

42 words · 1 min read

If $f$ is continuous on $[a, b]$ and $F' = f$ , then $\int_a^b f(x)\,dx = F(b) - F(a) = \Big[F(x)\Big]_a^b$ . No $+C$ is needed. The function must be continuous on the whole interval.