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In this section, we will learn:

To substitute a new variable in place of an existing expression in a function, making integration easier.

To substitute a new variable in place of an existing expression in a function, making integration easier.

1.
5

2.
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.

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.

3.
Due to the Fundamental Theorem

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

able to find antiderivatives.

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

able to find antiderivatives.

4.
INTRODUCTION Equation 1

However, our antidifferentiation formulas

don’t tell us how to evaluate integrals such

2

2 x 1 x dx

However, our antidifferentiation formulas

don’t tell us how to evaluate integrals such

2

2 x 1 x dx

5.
To find this integral, we use the problem-

solving strategy of introducing something

The ‘something extra’ is a new variable.

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

solving strategy of introducing something

The ‘something extra’ is a new variable.

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

6.
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.

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

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

7.
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.

an integral were to be interpreted as

a differential, then the differential 2x dx

would occur in Equation 1.

8.
INTRODUCTION Equation 2

So, formally, without justifying our calculation,

we could write:

2 2

2 x 1 x dx 1 x 2 x dx

udu

3/ 2

u

2

3 C

2 3/ 2

( x 1)

2

3 C

So, formally, without justifying our calculation,

we could write:

2 2

2 x 1 x dx 1 x 2 x dx

udu

3/ 2

u

2

3 C

2 3/ 2

( x 1)

2

3 C

9.
However, now we can check that we have

the correct answer by using the Chain Rule

to differentiate the final function of Equation 2:

d 2 2

3 ( x 1)3 2 C 23 23 ( x 2 1)1 2 2 x

dx

2

2 x x 1

the correct answer by using the Chain Rule

to differentiate the final function of Equation 2:

d 2 2

3 ( x 1)3 2 C 23 23 ( x 2 1)1 2 2 x

dx

2

2 x x 1

10.
In general, this method works whenever

we have an integral that we can write in

the form

∫ f(g(x))g’(x) dx

we have an integral that we can write in

the form

∫ f(g(x))g’(x) dx

11.
INTRODUCTION Equation 3

Observe that, if F’ = f, then

∫ F’(g(x))g’(x) dx = F(g(x)) + C

because, by the Chain Rule,

d

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

dx

Observe that, if F’ = f, then

∫ F’(g(x))g’(x) dx = F(g(x)) + C

because, by the Chain Rule,

d

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

dx

12.
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

‘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

13.
Writing F’ = f, we get:

∫ f(g(x))g’(x) dx = ∫ f(u) du

Thus, we have proved the following rule.

∫ f(g(x))g’(x) dx = ∫ f(u) du

Thus, we have proved the following rule.

14.
SUBSTITUTION RULE Equation 4

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

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

15.
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.

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.

16.
SUBSTITUTION RULE

Thus, the Substitution Rule says:

It is permissible to operate with

dx and du after integral signs as if

they were differentials.

Thus, the Substitution Rule says:

It is permissible to operate with

dx and du after integral signs as if

they were differentials.

17.
SUBSTITUTION RULE Example 1

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.

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.

18.
SUBSTITUTION RULE Example 1

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

Rule, we have:

3 4

x cos( x 2) dx cos

4 u 1

du 1

4 cos u du

14 sin u C

4

sin( x 2) C

1

4

Notice that, at the final stage, we had to return to

the original variable x.

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

Rule, we have:

3 4

x cos( x 2) dx cos

4 u 1

du 1

4 cos u du

14 sin u C

4

sin( x 2) C

1

4

Notice that, at the final stage, we had to return to

the original variable x.

19.
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.

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.

20.
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.

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.

21.
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

If that is not possible, try choosing u to be

some complicated part of the integrand—

perhaps the inner function in a composite

22.
SUBSTITUTION RULE

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.

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.

23.
SUBSTITUTION RULE E. g. 2—Solution 1

Evaluate 2 x 1 dx

Let u = 2x + 1.

Then, du = 2 dx.

So, dx = du/2.

Evaluate 2 x 1 dx

Let u = 2x + 1.

Then, du = 2 dx.

So, dx = du/2.

24.
SUBSTITUTION RULE E. g. 2—Solution 1

Thus, the rule gives:

du

2 x 1 dx u

2

12

1

2 u du

32

u

1

2 C

3/ 2

1 32

3 u C

32

(2 x 1)

1

3 C

Thus, the rule gives:

du

2 x 1 dx u

2

12

1

2 u du

32

u

1

2 C

3/ 2

1 32

3 u C

32

(2 x 1)

1

3 C

25.
SUBSTITUTION RULE E. g. 2—Solution 2

Another possible substitution is u 2 x 1

dx

Then, du

2 x 1

So, dx 2 x 1

Alternatively, observe that u2 = 2x + 1.

So, 2u du = 2 dx.

Another possible substitution is u 2 x 1

dx

Then, du

2 x 1

So, dx 2 x 1

Alternatively, observe that u2 = 2x + 1.

So, 2u du = 2 dx.

26.
SUBSTITUTION RULE E. g. 2—Solution 2

Thus, 2 x 1 dx u u du

2

u du

3

u

C

3

32

3 (2 x 1) C

1

Thus, 2 x 1 dx u u du

2

u du

3

u

C

3

32

3 (2 x 1) C

1

27.
SUBSTITUTION RULE Example 3

x

Find 1 4x dx

2

Let u = 1 – 4x2.

Then, du = -8x dx.

So, x dx = -1/8 du and

x 1 1 2

dx du

1

8

1

8 du

u

1 4 x2 u

18 (2 u ) C 1

4 1 4x2 C

x

Find 1 4x dx

2

Let u = 1 – 4x2.

Then, du = -8x dx.

So, x dx = -1/8 du and

x 1 1 2

dx du

1

8

1

8 du

u

1 4 x2 u

18 (2 u ) C 1

4 1 4x2 C

28.
SUBSTITUTION RULE

The answer to the example could be

checked by differentiation.

Instead, let’s check it with a graph.

The answer to the example could be

checked by differentiation.

Instead, let’s check it with a graph.

29.
SUBSTITUTION RULE

Here, we have used a computer to graph

2

both the integrand f ( x ) x / 1 4 x

2

and its indefinite integral g ( x) 4 1 4 x

1

We take the case

C = 0.

Here, we have used a computer to graph

2

both the integrand f ( x ) x / 1 4 x

2

and its indefinite integral g ( x) 4 1 4 x

1

We take the case

C = 0.

30.
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

Notice that g(x):

Decreases when f(x) is negative

Increases when f(x) is positive

Has its minimum value when f(x) = 0

31.
SUBSTITUTION RULE

So, it seems reasonable, from the graphical

evidence, that g is an antiderivative of f.

So, it seems reasonable, from the graphical

evidence, that g is an antiderivative of f.

32.
SUBSTITUTION RULE Example 4

Calculate ∫ e5x dx

If we let u = 5x, then du = 5 dx.

So, dx = 1/5 du.

Therefore, 5x u

e dx e du

5

1

u

e C

1

5

5x

e C

1

5

Calculate ∫ e5x dx

If we let u = 5x, then du = 5 dx.

So, dx = 1/5 du.

Therefore, 5x u

e dx e du

5

1

u

e C

1

5

5x

e C

1

5

33.
SUBSTITUTION RULE Example 5

2 5

Find 1 x x dx

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 5

Find 1 x x dx

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.

34.
SUBSTITUTION RULE Example 5

Also, x2 = u – 1; so, x4 = (u – 1)2:

2 5 2 4 du

2

1 x x dx 1 x x x dx u (u 1)

2

12 u (u 2 2u 1) du

5/ 2 3/ 2 1/ 2

1

2 (u 2u u ) du

7/2 5/ 2 3/ 2

( u

1

2

2

7 2 u 2

5 u2

3 ) C

17 (1 x 2 )7 / 2 52 (1 x 2 )5/ 2

13 (1 x 2 )3/ 2 C

Also, x2 = u – 1; so, x4 = (u – 1)2:

2 5 2 4 du

2

1 x x dx 1 x x x dx u (u 1)

2

12 u (u 2 2u 1) du

5/ 2 3/ 2 1/ 2

1

2 (u 2u u ) du

7/2 5/ 2 3/ 2

( u

1

2

2

7 2 u 2

5 u2

3 ) C

17 (1 x 2 )7 / 2 52 (1 x 2 )5/ 2

13 (1 x 2 )3/ 2 C

35.
SUBSTITUTION RULE Example 6

Calculate ∫ tan x dx

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

sin x

tan x dx cos x dx

This suggests that we should substitute u = cos x,

since then du = – sin x dx, and so sin x dx = – du:

sin x du

tan x dx cos x dx u

ln | u | C

ln | cos x | C

Calculate ∫ tan x dx

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

sin x

tan x dx cos x dx

This suggests that we should substitute u = cos x,

since then du = – sin x dx, and so sin x dx = – du:

sin x du

tan x dx cos x dx u

ln | u | C

ln | cos x | C

36.
SUBSTITUTION RULE Equation 5

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

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

37.
DEFINITE INTEGRALS

When evaluating a definite integral

by substitution, two methods are

When evaluating a definite integral

by substitution, two methods are

38.
DEFINITE INTEGRALS

One method is to evaluate the indefinite

integral first and then use the FTC.

For instance, using the result of Example 2,

we have: 4 4

2 x 1 dx 2 x 1 dx

0 0

32 4

(2 x 1)

1

3 0

32 32

(9)

1

3 (1)

1

3

13 (27 1) 263

One method is to evaluate the indefinite

integral first and then use the FTC.

For instance, using the result of Example 2,

we have: 4 4

2 x 1 dx 2 x 1 dx

0 0

32 4

(2 x 1)

1

3 0

32 32

(9)

1

3 (1)

1

3

13 (27 1) 263

39.
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.

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.

40.
SUB. RULE FOR DEF. INTEGRALS Equation 6

If g’ is continuous on [a, b] and f

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

b g (b )

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

a g (a)

f (u )du

If g’ is continuous on [a, b] and f

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

b g (b )

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

a g (a)

f (u )du

41.
SUB. RULE FOR DEF. INTEGRALS Proof

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

f ( g ( x)) g '( x)dx F ( g ( x))

a a

F ( g (b)) F ( g (a ))

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

f ( g ( x)) g '( x)dx F ( g ( x))

a a

F ( g (b)) F ( g (a ))

42.
SUB. RULE FOR DEF. INTEGRALS Proof

However, applying the FTC2 a second time,

we also have:

g (b ) g (b )

f (u ) du F (u ) g ( a )

g (a)

F ( g (b)) F ( g (a ))

However, applying the FTC2 a second time,

we also have:

g (b ) g (b )

f (u ) du F (u ) g ( a )

g (a)

F ( g (b)) F ( g (a ))

43.
SUB. RULE FOR DEF. INTEGRALS Example 7

4

Evaluate 2 x 1 dx using Equation 6.

0

Using the substitution

from Solution 1 of

Example 2, we have:

u = 2x + 1 and dx = du/2

4

Evaluate 2 x 1 dx using Equation 6.

0

Using the substitution

from Solution 1 of

Example 2, we have:

u = 2x + 1 and dx = du/2

44.
SUB. RULE FOR DEF. INTEGRALS Example 7

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

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

45.
SUB. RULE FOR DEF. INTEGRALS Example 7

4 9

Thus, 2 x 1 dx 12 u du

0 1

32 9

12 23 u

1

32 32

(91

3 1 )

26

3

4 9

Thus, 2 x 1 dx 12 u du

0 1

32 9

12 23 u

1

32 32

(91

3 1 )

26

3

46.
SUB. RULE FOR DEF. INTEGRALS Example 7

Observe that, when using Equation 6,

we do not return to the variable x after

We simply evaluate the expression in u

between the appropriate values of u.

Observe that, when using Equation 6,

we do not return to the variable x after

We simply evaluate the expression in u

between the appropriate values of u.

47.
SUB. RULE FOR DEF. INTEGRALS Example 8

2 dx

Evaluate

1 (3 5 x ) 2

Let u = 3 - 5x.

Then, du = – 5 dx, so dx = – du/5.

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

2 dx

Evaluate

1 (3 5 x ) 2

Let u = 3 - 5x.

Then, du = – 5 dx, so dx = – du/5.

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

48.
SUB. RULE FOR DEF. INTEGRALS Example 8

2 dx 1 7 du

Thus, 2

2

1 (3 5 x ) 5 2 u

7

1 1

5 u 5

7

1

5u 2

1 1 1 1

5 7 2 14

2 dx 1 7 du

Thus, 2

2

1 (3 5 x ) 5 2 u

7

1 1

5 u 5

7

1

5u 2

1 1 1 1

5 7 2 14

49.
SUB. RULE FOR DEF. INTEGRALS Example 9

ln xe

Calculate dx

1 x

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.

2 1

ln

e x 1 u 1

Thus,

1 x dx 0 u du 2 2

0

ln xe

Calculate dx

1 x

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.

2 1

ln

e x 1 u 1

Thus,

1 x dx 0 u du 2 2

0

50.
SUB. RULE FOR DEF. INTEGRALS Example 9

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.

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.

51.
The next theorem uses the Substitution

Rule for Definite Integrals to simplify

the calculation of integrals of functions that

possess symmetry properties.

Rule for Definite Integrals to simplify

the calculation of integrals of functions that

possess symmetry properties.

52.
INTEGS. OF SYMM. FUNCTIONS Theorem 7

Suppose f is continuous on [–a , a].

a.If f is even, [f(–x) = f(x)], then

a a

f ( x) dx 2 f ( x) dx

a 0

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

a

f ( x) dx 0

a

Suppose f is continuous on [–a , a].

a.If f is even, [f(–x) = f(x)], then

a a

f ( x) dx 2 f ( x) dx

a 0

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

a

f ( x) dx 0

a

53.
INTEGS. OF SYMM. FUNCTIONS Proof—Equation 8

We split the integral in two:

a 0 a

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

a a 0

a a

f ( x) dx f ( x) dx

0 0

We split the integral in two:

a 0 a

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

a a 0

a a

f ( x) dx f ( x) dx

0 0

54.
INTEGS. OF SYMM. FUNCTIONS Proof

a 0 a

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

a a 0

a a

f ( x) dx f ( x) dx

0 0

In the first integral in the second part,

we make the substitution u = –x .

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

a 0 a

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

a a 0

a a

f ( x) dx f ( x) dx

0 0

In the first integral in the second part,

we make the substitution u = –x .

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

55.
INTEGS. OF SYMM. FUNCTIONS Proof

a a

f ( x) dx f ( u )( du )

0 0

a

f ( u ) du

0

a a

f ( x) dx f ( u )( du )

0 0

a

f ( u ) du

0

56.
INTEGS. OF SYMM. FUNCTIONS Proof—Equation 9

So, Equation 8 becomes:

a

f ( x) dx

a

a a

f ( u ) du f ( x) dx

0 0

So, Equation 8 becomes:

a

f ( x) dx

a

a a

f ( u ) du f ( x) dx

0 0

57.
INTEGS. OF SYMM. FUNCTIONS Proof a

If f is even, then f(–u) = f(u).

a

So, Equation 9 gives: f ( x) dx

a

a a

f (u ) du f ( x) dx

0 0

a

2 f ( x) dx

0

If f is even, then f(–u) = f(u).

a

So, Equation 9 gives: f ( x) dx

a

a a

f (u ) du f ( x) dx

0 0

a

2 f ( x) dx

0

58.
INTEGS. OF SYMM. FUNCTIONS Proof b

If f is odd, then f(–u) = –f(u).

a

So, Equation 9 gives: f ( x) dx

a

a a

f (u ) du f ( x) dx

0 0

0

If f is odd, then f(–u) = –f(u).

a

So, Equation 9 gives: f ( x) dx

a

a a

f (u ) du f ( x) dx

0 0

0

59.
INTEGS. OF SYMM. FUNCTIONS

Theorem 7 is

illustrated here.

Theorem 7 is

illustrated here.

60.
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.

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.

61.
INTEGS. OF SYMM. FUNCTIONS

b

Recall that an integral f ( x) dx can be

a

expressed as the area above the x–axis

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

the axis and above the curve.

b

Recall that an integral f ( x) dx can be

a

expressed as the area above the x–axis

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

the axis and above the curve.

62.
INTEGS. OF SYMM. FUNCTIONS

Therefore, part (b) says the integral

is 0 because the areas cancel.

Therefore, part (b) says the integral

is 0 because the areas cancel.

63.
INTEGS. OF SYMM. FUNCTIONS Example 10

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

2 2

6 6

So, (x

2

1) dx 2 ( x 1) dx

0

7 2

2 x x

1

7 0

2 128

7 2

284

7

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

2 2

6 6

So, (x

2

1) dx 2 ( x 1) dx

0

7 2

2 x x

1

7 0

2 128

7 2

284

7

64.
INTEGS. OF SYMM. FUNCTIONS Example 11

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

f(–x) = –f(x), it is odd.

1 tan x

So, 2 4

dx 0

11 x x

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

f(–x) = –f(x), it is odd.

1 tan x

So, 2 4

dx 0

11 x x