INTEGRALS 5. 5.5 The Substitution Rule In this section, we will learn: To substitute a new variable in place of an existing expression in a function,

INTEGRALSINTEGRALS

5

5.5The Substitution Rule

In this section, we will learn:

To substitute a new variable in place of an existing

expression in a function, making integration easier.

INTEGRALS

Due to the Fundamental Theorem

of Calculus (FTC), it’s important to be

able to find antiderivatives.

INTRODUCTION

However, our antidifferentiation formulas

don’t tell us how to evaluate integrals such

as22 1x x dx

Equation 1INTRODUCTION

To find this integral, we use the problem-

solving strategy of introducing something

extra.

The ‘something extra’ is a new variable.

We change from the variable x to a new variable u.

INTRODUCTION

Suppose we let u be the quantity under

the root sign in Equation 1, u = 1 + x2.

Then, the differential of u is du = 2x dx.

INTRODUCTION

Notice that, if the dx in the notation for

an integral were to be interpreted as

a differential, then the differential 2x dx

would occur in Equation 1.

INTRODUCTION

So, formally, without justifying our calculation,

we could write:

2 2

3/ 223

2 3/ 223

2 1 1 2

( 1)

x x dx x x dx

udu

u C

x C

Equation 2INTRODUCTION

However, now we can check that we have

the correct answer by using the Chain Rule

to differentiate the final function of Equation 2:

2 3 2 2 1 232 23 3 2

2

( 1) ( 1) 2

2 1

dx C x x

dx

x x

INTRODUCTION

In general, this method works whenever

we have an integral that we can write in

the form

∫ f(g(x))g’(x) dx

INTRODUCTION

Observe that, if F’ = f, then

∫ F’(g(x))g’(x) dx = F(g(x)) + C

because, by the Chain Rule,

Equation 3INTRODUCTION

( ( )) '( ( )) '( )dF g x F g x g x

dx

If we make the ‘change of variable’ or

‘substitution’ u = g(x), from Equation 3,

we have: '( ( )) '( ) ( ( ))

( )

'( )

F g x g x dx F g x C

F u C

F u du

INTRODUCTION

Writing F’ = f, we get:

∫ f(g(x))g’(x) dx = ∫ f(u) du

Thus, we have proved the following rule.

INTRODUCTION

SUBSTITUTION RULE

If u = g(x) is a differentiable function whose

range is an interval I and f is continuous

on I, then

∫ f(g(x))g’(x) dx = ∫ f(u) du

Equation 4

SUBSTITUTION RULE

Notice that the Substitution Rule for

integration was proved using the Chain Rule

for differentiation.

Notice also that, if u = g(x), then du = g’(x) dx.

So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.

SUBSTITUTION RULE

Thus, the Substitution Rule says:

It is permissible to operate with

dx and du after integral signs as if

they were differentials.

SUBSTITUTION RULE

Find ∫ x3 cos(x4 + 2) dx

We make the substitution u = x4 + 2.

This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.

Example 1

SUBSTITUTION RULE

Thus, using x3 dx = du/4 and the Substitution

Rule, we have:

Notice that, at the final stage, we had to return to the original variable x.

3 4 1 14 4

14

414

cos( 2) cos cos

sin

sin( 2)

x x dx u du u du

u C

x C

Example 1

SUBSTITUTION RULE

The idea behind the Substitution Rule is to

replace a relatively complicated integral by

a simpler integral.

This is accomplished by changing from the original variable x to a new variable u that is a function of x.

Thus, in Example 1, we replaced the integral ∫ x3cos(x4 + 2) dx by the simpler integral ¼ ∫ cos u du.

SUBSTITUTION RULE

The main challenge in using the rule is

to think of an appropriate substitution.

You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor.

This was the case in Example 1.

SUBSTITUTION RULE

If that is not possible, try choosing u to be

some complicated part of the integrand—

perhaps the inner function in a composite

function.

Finding the right substitution is

a bit of an art.

It’s not unusual to guess wrong.

If your first guess doesn’t work, try another substitution.

SUBSTITUTION RULE

SUBSTITUTION RULE

Evaluate

Let u = 2x + 1.

Then, du = 2 dx.

So, dx = du/2.

2 1x dxE. g. 2—Solution 1

Thus, the rule gives:

1 212

3 212

3 213

3 213

2 12

3/ 2

(2 1)

dux dx u

u du

uC

u C

x C

SUBSTITUTION RULE E. g. 2—Solution 1

Another possible substitution is

Then,

So,

Alternatively, observe that u2 = 2x + 1. So, 2u du = 2 dx.

SUBSTITUTION RULE E. g. 2—Solution 2

2 1u x

2 1

dxdu

x

2 1dx x

Thus,

SUBSTITUTION RULE E. g. 2—Solution 2

2

3

3 213

2 1

3

(2 1)

x dx u u du

u du

uC

x C

SUBSTITUTION RULE

Find

Let u = 1 – 4x2. Then, du = -8x dx. So, x dx = -1/8 du and

21 4

xdx

x

1 21 18 82

21 18 4

1

1 4

(2 ) 1 4

xdx du u du

ux

u C x C

Example 3

SUBSTITUTION RULE

The answer to the example could be

checked by differentiation.

Instead, let’s check it with a graph.

SUBSTITUTION RULE

Here, we have used a computer to graph

both the integrand

and its indefinite integral

We take the case C = 0.

2( ) / 1 4f x x x 21

4( ) 1 4g x x

SUBSTITUTION RULE

Notice that g(x):

Decreases when f(x) is negative Increases when f(x) is positive Has its minimum value when f(x) = 0

So, it seems reasonable, from the graphical

evidence, that g is an antiderivative of f.

SUBSTITUTION RULE

SUBSTITUTION RULE

Calculate ∫ e5x dx

If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore, 5 1

5

15

515

x u

u

x

e dx e du

e C

e C

Example 4

SUBSTITUTION RULE

Find

An appropriate substitution becomes more obvious if we factor x5 as x4 . x.

Let u = 1 + x2.

Then, du = 2x dx.

So, x dx = du/2.

2 51 x x dxExample 5

SUBSTITUTION RULE

Also, x2 = u – 1; so, x4 = (u – 1)2:

2 5 2 4 2

212

5/ 2 3/ 2 1/ 212

7 / 2 5/ 2 3/ 21 2 2 22 7 5 3

2 7 / 2 2 5/ 21 27 5

2 3/ 213

1 1 ( 1)2

( 2 1)

( 2 )

( 2 )

(1 ) (1 )

(1 )

dux x dx x x x dx u u

u u u du

u u u du

u u u C

x x

x C

Example 5

SUBSTITUTION RULE

Calculate ∫ tan x dx

First, we write tangent in terms of sine and cosine:

This suggests that we should substitute u = cos x, since then du = – sin x dx, and so sin x dx = – du:

sintan

cos

xx dx dx

x

sintan

cosln | |

ln | cos |

x dux dx dx

x uu C

x C

Example 6

SUBSTITUTION RULE

Since –ln |cos x| = ln(|cos x|-1)

= ln(1/|cos x|)

= ln|sec x|,

the result of the example can also be written

as ∫ tan x dx = ln |sec x| + C

Equation 5

DEFINITE INTEGRALS

When evaluating a definite integral

by substitution, two methods are

possible.

One method is to evaluate the indefinite

integral first and then use the FTC.

For instance, using the result of Example 2, we have:

DEFINITE INTEGRALS

44

0 0

43 213 0

3 2 3 21 13 3

2613 3

2 1 2 1

(2 1)

(9) (1)

(27 1)

x dx x dx

x

DEFINITE INTEGRALS

Another method, which is usually preferable,

is to change the limits of integration when

the variable is changed.

Thus, we have the substitution rule for

definite integrals.

If g’ is continuous on [a, b] and f

is continuous on the range of u = g(x),

then

( )

( )( ( )) '( ) ( )

b g b

a g af g x g x dx f u du

SUB. RULE FOR DEF. INTEGRALS Equation 6

Let F be an antiderivative of f.

Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x).

So, by Part 2 of the FTC (FTC2), we have:

( ( )) '( ) ( ( ))

( ( )) ( ( ))

b b

aaf g x g x dx F g x

F g b F g a

SUB. RULE FOR DEF. INTEGRALS Proof

However, applying the FTC2 a second time, we also have:

( ) ( )

( )( )( ) ( )

( ( )) ( ( ))

g b g b

g ag af u du F u

F g b F g a

SUB. RULE FOR DEF. INTEGRALS Proof

Evaluate using Equation 6.

Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2

4

02 1x dx

Example 7SUB. RULE FOR DEF. INTEGRALS

To find the new limits of integration, we

note that: When x = 0, u = 2(0) + 1 = 1 When x = 4, u = 2(4) + 1 = 9

Example 7SUB. RULE FOR DEF. INTEGRALS

Thus,4 9

120 1

93 21 22 3 1

3 2 3 213

263

2 1

(9 1 )

x dx u du

u

Example 7SUB. RULE FOR DEF. INTEGRALS

Observe that, when using Equation 6,

we do not return to the variable x after

integrating.

We simply evaluate the expression in u between the appropriate values of u.

SUB. RULE FOR DEF. INTEGRALS Example 7

Evaluate

Let u = 3 - 5x.

Then, du = – 5 dx, so dx = – du/5.

When x = 1, u = – 2, and when x = 2, u = – 7.

2

21 (3 5 )

dx

xExample 8SUB. RULE FOR DEF. INTEGRALS

Thus,2 7

2 21 2

7

5

7

2

1

(3 5 ) 5

1 1

5

1

5

1 1 1 1

5 7 2 14

dx du

x u

u

u

Example 8SUB. RULE FOR DEF. INTEGRALS

Calculate

We let u = ln x because its differential du = dx/x occurs in the integral.

When x = 1, u = ln 1, and when x = e, u = ln e = 1.

Thus,

1

lne xdx

xExample 9SUB. RULE FOR DEF. INTEGRALS

121

1 00

ln 1

2 2

e x udx u du

x

As the function f(x) = (ln x)/x in the example

is positive for x > 1, the integral represents

the area of the shaded region in this figure.

SUB. RULE FOR DEF. INTEGRALS Example 9

SYMMETRY

The next theorem uses the Substitution

Rule for Definite Integrals to simplify

the calculation of integrals of functions that

possess symmetry properties.

INTEGS. OF SYMM. FUNCTIONS

Suppose f is continuous on [–a , a].

a.If f is even, [f(–x) = f(x)], then

b.If f is odd, [f(-x) = -f(x)], then

0( ) 2 ( )

a a

af x dx f x dx

( ) 0a

af x dx

Theorem 7

We split the integral in two:

0

0

0 0

( ) ( ) ( )

( ) ( )

a a

a a

a a

f x dx f x dx f x dx

f x dx f x dx

Proof—Equation 8INTEGS. OF SYMM. FUNCTIONS

In the first integral in the second part,

we make the substitution u = –x .

Then, du = –dx, and when x = –a, u = a.

INTEGS. OF SYMM. FUNCTIONS0

0

0 0

( ) ( ) ( )

( ) ( )

a a

a a

a a

f x dx f x dx f x dx

f x dx f x dx

Proof

Therefore,

0 0

0

( ) ( )( )

( )

a a

a

f x dx f u du

f u du

ProofINTEGS. OF SYMM. FUNCTIONS

So, Equation 8 becomes:

0 0

( )

( ) ( )

a

a

a a

f x dx

f u du f x dx

Proof—Equation 9INTEGS. OF SYMM. FUNCTIONS

If f is even, then f(–u) = f(u).

So, Equation 9 gives:

0 0

0

( )

( ) ( )

2 ( )

a

a

a a

a

f x dx

f u du f x dx

f x dx

INTEGS. OF SYMM. FUNCTIONS Proof a

If f is odd, then f(–u) = –f(u).

So, Equation 9 gives:

INTEGS. OF SYMM. FUNCTIONS Proof b

0 0

( )

( ) ( )

0

a

a

a a

f x dx

f u du f x dx

Theorem 7 is

illustrated here.

INTEGS. OF SYMM. FUNCTIONS

For the case where

f is positive and

even, part (a) says

that the area under

y = f(x) from –a to a

is twice the area

from 0 to a because

of symmetry.

INTEGS. OF SYMM. FUNCTIONS

Recall that an integral can be

expressed as the area above the x–axis

and below y = f(x) minus the area below

the axis and above the curve.

INTEGS. OF SYMM. FUNCTIONS

( )b

af x dx

Therefore, part (b) says the integral

is 0 because the areas cancel.

INTEGS. OF SYMM. FUNCTIONS

As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even.

So,

2 26 6

2 0

2717 0

1287

2847

( 1) 2 ( 1)

2

2 2

x dx x dx

x x

Example 10INTEGS. OF SYMM. FUNCTIONS

As f(x) = (tan x)/ (1 + x2 + x4) satisfies

f(–x) = –f(x), it is odd.

So, 1

2 41

tan0

1

xdx

x x

Example 11INTEGS. OF SYMM. FUNCTIONS