5.5 The Substitution Rule In this section, we will learn: To substitute a new variable in place of...

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, making integration easier.

INTEGRALS

Due to the Fundamental Theorem of Calculus, it is

important to be able to find antiderivatives.

However, our antidifferentiation formulas do not

tell us how to evaluate integrals such as

INTRODUCTION

22 1x x dx Equation 1

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. 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

INTRODUCTION

( ( )) ( )f g x g x dx

Observe that, if F’ = f, then

because, by the Chain Rule,

Equation 3INTRODUCTION

( ( )) '( ( )) '( )d

F g x F g x g xdx

( ( )) ( ) ( ( ))F g x g x dx F g x C

That is, 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:

Thus, we have proved the following rule.

INTRODUCTION

( ( )) '( ) ( )f g x g x dx f u du

SUBSTITUTION RULE

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

is an interval I and f is continuous on I, then

Equation 4

( ( )) '( ) ( )f g x g x dx f u du

SUBSTITUTION RULE

Notice that the Substitution Rule 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

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

3 4cos( 2)x x dx

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

by the simpler integral3 4cos( 2)x x dx1

cos4

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 is not unusual to guess wrong.

If your first guess does not work, try another substitution.

SUBSTITUTION RULE

SUBSTITUTION RULE

Evaluate

Let u = 2x + 1.

Then, du = 2 dx.

So, dx = du/2.

2 1x dx

E. 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 du udu

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, xdx = -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 us 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

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

5 15

15

515

x u

u

x

e dx e du

e C

e C

Example 4

5xe dx

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.

5 21x x dx

Example 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

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 ln | |

cosln | cos |

x dux dx dx u C

x ux C

Example 6

tan x dx

SUBSTITUTION RULE

Since –ln|cos x| = ln(|cos x|-1)

= ln(1/|cos x|)

= ln|sec x|,

the result can also be written as

Equation 5

tan ln | sec |x dx x C

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 and when x = 4, u = 2(4)

+ 1 = 9

Example 7SUB. RULE FOR DEF. INTEGRALS

Thus,

4 9120 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

x

Example 8SUB. RULE FOR DEF. INTEGRALS

Thus,2 7

2 21 2

7

2

1

(3 5 ) 5

1 1

5

1 1 1 1

5 7 2 14

dx du

x 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

x

Example 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. FUNCTIONS

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

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),

thus f(x) is odd and,

1

2 41

tan0

1

xdx

x x

Example 11INTEGS. OF SYMM. FUNCTIONS