Contributed by:

In this section, we will learn about:

The techniques to evaluate miscellaneous integrals.

The techniques to evaluate miscellaneous integrals.

1.
7

TECHNIQUES OF INTEGRATION

TECHNIQUES OF INTEGRATION

2.
TECHNIQUES OF INTEGRATION

As we have seen, integration is more

challenging than differentiation.

In finding the derivative of a function, it is obvious

which differentiation formula we should apply.

However, it may not be obvious which technique

we should use to integrate a given function.

As we have seen, integration is more

challenging than differentiation.

In finding the derivative of a function, it is obvious

which differentiation formula we should apply.

However, it may not be obvious which technique

we should use to integrate a given function.

3.
TECHNIQUES OF INTEGRATION

Until now, individual techniques have been

applied in each section.

For instance, we usually used:

Substitution in Exercises 5.5

Integration by parts in Exercises 7.1

Partial fractions in Exercises 7.4

Until now, individual techniques have been

applied in each section.

For instance, we usually used:

Substitution in Exercises 5.5

Integration by parts in Exercises 7.1

Partial fractions in Exercises 7.4

4.
TECHNIQUES OF INTEGRATION

7.5

Strategy for Integration

In this section, we will learn about:

The techniques to evaluate miscellaneous integrals.

7.5

Strategy for Integration

In this section, we will learn about:

The techniques to evaluate miscellaneous integrals.

5.
STRATEGY FOR INTEGRATION

In this section, we present a collection

of miscellaneous integrals in random order.

The main challenge is to recognize which

technique or formula to use.

In this section, we present a collection

of miscellaneous integrals in random order.

The main challenge is to recognize which

technique or formula to use.

6.
STRATEGY FOR INTEGRATION

No hard and fast rules can be given as to

which method applies in a given situation.

However, we give some advice on strategy

that you may find useful.

No hard and fast rules can be given as to

which method applies in a given situation.

However, we give some advice on strategy

that you may find useful.

7.
STRATEGY FOR INTEGRATION

A prerequisite for strategy selection

is a knowledge of the basic integration

A prerequisite for strategy selection

is a knowledge of the basic integration

8.
STRATEGY FOR INTEGRATION

In the upcoming table, we have

The integrals from our previous list

Several additional formulas we have learned

in this chapter

In the upcoming table, we have

The integrals from our previous list

Several additional formulas we have learned

in this chapter

9.
STRATEGY FOR INTEGRATION

Most should be memorized.

It is useful to know them all.

However, the ones marked with an asterisk

need not be memorized—they are easily derived.

Most should be memorized.

It is useful to know them all.

However, the ones marked with an asterisk

need not be memorized—they are easily derived.

10.
TABLE OF INTEGRATION FORMULAS

n 1

n x 1

1. x dx (n 1) 2. dx ln | x |

n 1 x

x

x x x a

3. e dx e 4. a dx

ln a

n 1

n x 1

1. x dx (n 1) 2. dx ln | x |

n 1 x

x

x x x a

3. e dx e 4. a dx

ln a

11.
TABLE OF INTEGRATION FORMULAS

5. sin x dx cos x 6.cos x dx sin x

7. sec 2 x dx tan x 8.csc 2 x dx cot x

9.sec x tan x dx sec x 10.csc x cot x dx csc x

11. sec x dx ln sec x tan x 12. csc x dx ln csc x cot x

5. sin x dx cos x 6.cos x dx sin x

7. sec 2 x dx tan x 8.csc 2 x dx cot x

9.sec x tan x dx sec x 10.csc x cot x dx csc x

11. sec x dx ln sec x tan x 12. csc x dx ln csc x cot x

12.
TABLE OF INTEGRATION FORMULAS

13. tan x dx ln sec x 14. cot x dx ln sin x

15.sinh x dx cosh x 16.cosh x dx sinh x

dx 1 1 x dx x

1

17. 2 2

tan 18. sin

x a a a 2

a x 2

a

13. tan x dx ln sec x 14. cot x dx ln sin x

15.sinh x dx cosh x 16.cosh x dx sinh x

dx 1 1 x dx x

1

17. 2 2

tan 18. sin

x a a a 2

a x 2

a

13.
TABLE OF INTEGRATION FORMULAS

dx 1 x a dx

*19. 2 2

ln *20. ln x x 2 a 2

x a 2a x a x 2 a 2

Formula 19 can be avoided by using partial fractions.

Trigonometric substitutions can be used instead of

Formula 20.

dx 1 x a dx

*19. 2 2

ln *20. ln x x 2 a 2

x a 2a x a x 2 a 2

Formula 19 can be avoided by using partial fractions.

Trigonometric substitutions can be used instead of

Formula 20.

14.
STRATEGY FOR INTEGRATION

Once armed with these basic integration

formulas, you might try this strategy:

1. Simplify the integrand if possible.

2. Look for an obvious substitution.

3. Classify the integrand according to its form.

4. Try again.

Once armed with these basic integration

formulas, you might try this strategy:

1. Simplify the integrand if possible.

2. Look for an obvious substitution.

3. Classify the integrand according to its form.

4. Try again.

15.
1. SIMPLIFY THE INTEGRAND

Sometimes, the use of algebraic

manipulation or trigonometric identities

will simplify the integrand and make

the method of integration obvious.

Sometimes, the use of algebraic

manipulation or trigonometric identities

will simplify the integrand and make

the method of integration obvious.

16.
1. SIMPLIFY THE INTEGRAND

Here are some examples:

x 1 x dx

x x dx

tan sin 2

sec2 d cos cos d sin cos d 1

2sin 2 d

2 2 2

(sin x cos x ) dx (sin x 2sin x cos x cos x) dx

(1 2sin x cos x) dx

Here are some examples:

x 1 x dx

x x dx

tan sin 2

sec2 d cos cos d sin cos d 1

2sin 2 d

2 2 2

(sin x cos x ) dx (sin x 2sin x cos x cos x) dx

(1 2sin x cos x) dx

17.
2. LOOK FOR OBVIOUS SUBSTITUTION

Try to find some function u = g(x) in

the integrand whose differential du = g’(x) dx

also occurs, apart from a constant factor.

x

For instance, in the integral x 2 1 dx ,

notice that, if u = x2 – 1, then du = 2x dx.

So, we use the substitution u = x2 – 1

instead of the method of partial fractions.

Try to find some function u = g(x) in

the integrand whose differential du = g’(x) dx

also occurs, apart from a constant factor.

x

For instance, in the integral x 2 1 dx ,

notice that, if u = x2 – 1, then du = 2x dx.

So, we use the substitution u = x2 – 1

instead of the method of partial fractions.

18.
3. CLASSIFY THE FORM

If Steps 1 and 2 have not led to

the solution, we take a look at the form

of the integrand f(x).

If Steps 1 and 2 have not led to

the solution, we take a look at the form

of the integrand f(x).

19.
3 a. TRIGONOMETRIC FUNCTIONS

We use the substitutions recommended

in Section 7.2 if f(x) is a product of:

sin x and cos x

tan x and sec x

cot x and csc x

We use the substitutions recommended

in Section 7.2 if f(x) is a product of:

sin x and cos x

tan x and sec x

cot x and csc x

20.
3 b. RATIONAL FUNCTIONS

If f is a rational function, we

use the procedure of Section 7.4

involving partial fractions.

If f is a rational function, we

use the procedure of Section 7.4

involving partial fractions.

21.
3 c. INTEGRATION BY PARTS

If f(x) is a product of a power of x (or

a polynomial) and a transcendental function

(a trigonometric, exponential, or logarithmic

function), we try integration by parts.

We choose u and dv as per the advice in Section 7.1

If you look at the functions in Exercises 7.1, you

will see most of them are the type just described.

If f(x) is a product of a power of x (or

a polynomial) and a transcendental function

(a trigonometric, exponential, or logarithmic

function), we try integration by parts.

We choose u and dv as per the advice in Section 7.1

If you look at the functions in Exercises 7.1, you

will see most of them are the type just described.

22.
3 d. RADICALS

Particular kinds of substitutions are

recommended when certain radicals appear.

2 2

i. If x a occurs, we use a trigonometric

substitution according to the table in section 7.3

ii. If n ax b occurs, we use the rationalizing

substitution u n ax b .

More generally, this sometimes works for n g ( x) .

Particular kinds of substitutions are

recommended when certain radicals appear.

2 2

i. If x a occurs, we use a trigonometric

substitution according to the table in section 7.3

ii. If n ax b occurs, we use the rationalizing

substitution u n ax b .

More generally, this sometimes works for n g ( x) .

23.
4. TRY AGAIN

If the first three steps have not produced

the answer, remember there are basically

only two methods of integration:

1. Substitution

2. Parts

If the first three steps have not produced

the answer, remember there are basically

only two methods of integration:

1. Substitution

2. Parts

24.
4 a. TRY SUBSTITUTION

Even if no substitution is obvious (Step 2),

some inspiration or ingenuity (or even

desperation) may suggest an appropriate

Even if no substitution is obvious (Step 2),

some inspiration or ingenuity (or even

desperation) may suggest an appropriate

25.
4 b. TRY PARTS

Though integration by parts is used most

of the time on products of the form described

in Step 3 c, it is sometimes effective on single

Looking at Section 7.1, we see that it works on tan-1x,

sin-1x, and ln x, and these are all inverse functions.

Though integration by parts is used most

of the time on products of the form described

in Step 3 c, it is sometimes effective on single

Looking at Section 7.1, we see that it works on tan-1x,

sin-1x, and ln x, and these are all inverse functions.

26.
4 c. MANIPULATE THE INTEGRAND

Algebraic manipulations (rationalizing the

denominator, using trigonometric identities)

may be useful in transforming the integral

into an easier form.

These manipulations may be more substantial

than in Step 1 and may involve some ingenuity.

Algebraic manipulations (rationalizing the

denominator, using trigonometric identities)

may be useful in transforming the integral

into an easier form.

These manipulations may be more substantial

than in Step 1 and may involve some ingenuity.

27.
4 c. MANIPULATE THE INTEGRAND

Here is an example:

dx 1 1 cos x

1 cos x 1 cos x 1 cos x dx

1 cos x

2

dx

1 cos x

1 cos x

2 dx

sin x

2 cos x

csc x 2 dx

sin x

Here is an example:

dx 1 1 cos x

1 cos x 1 cos x 1 cos x dx

1 cos x

2

dx

1 cos x

1 cos x

2 dx

sin x

2 cos x

csc x 2 dx

sin x

28.
4 d. RELATE TO PREVIOUS PROBLEMS

When you have built up some experience in

integration, you may be able to use a method

on a given integral that is similar to a method

you have already used on a previous integral.

You may even be able to express the given

integral in terms of a previous one.

When you have built up some experience in

integration, you may be able to use a method

on a given integral that is similar to a method

you have already used on a previous integral.

You may even be able to express the given

integral in terms of a previous one.

29.
4 d. RELATE TO PREVIOUS PROBLEMS

For instance, ∫ tan2x sec x dx is

a challenging integral.

If we make use of the identity tan2x = sec2x – 1,

we can write:

2 3

tan x sec x dx sec x dx sec x dx

Then, if ∫ sec3x dx has previously been evaluated,

that calculation can be used in the present problem.

For instance, ∫ tan2x sec x dx is

a challenging integral.

If we make use of the identity tan2x = sec2x – 1,

we can write:

2 3

tan x sec x dx sec x dx sec x dx

Then, if ∫ sec3x dx has previously been evaluated,

that calculation can be used in the present problem.

30.
4 e. USE SEVERAL METHODS

Sometimes, two or three methods

are required to evaluate an integral.

The evaluation could involve several successive

substitutions of different types.

It might even combine integration by parts

with one or more substitutions.

Sometimes, two or three methods

are required to evaluate an integral.

The evaluation could involve several successive

substitutions of different types.

It might even combine integration by parts

with one or more substitutions.

31.
STRATEGY FOR INTEGRATION

In the following examples, we

indicate a method of attack, but

do not fully work out the integral.

In the following examples, we

indicate a method of attack, but

do not fully work out the integral.

32.
SIMPLIFY INTEGRAND Example 1

3

tan x

cos3 x dx

In Step 1, we rewrite the integral:

tan 3 x 3 3

cos3 x dx tan x sec x dx

It is now of the form ∫tanmx secnx dx with m odd.

So, we can use the advice in Section 7.2

3

tan x

cos3 x dx

In Step 1, we rewrite the integral:

tan 3 x 3 3

cos3 x dx tan x sec x dx

It is now of the form ∫tanmx secnx dx with m odd.

So, we can use the advice in Section 7.2

33.
TRY SUBSTITUTION Example 1

Suppose, in Step 1, we had written:

tan 3 x sin 3 x 1

cos3 x dx cos3 x cos3 x dx

3

sin x

6 dx

cos x

Then, we could have continued as follows.

Suppose, in Step 1, we had written:

tan 3 x sin 3 x 1

cos3 x dx cos3 x cos3 x dx

3

sin x

6 dx

cos x

Then, we could have continued as follows.

34.
TRY SUBSTITUTION Example 1

Substitute u = cos x:

sin 3 x 1 cos 2 x

cos6 x dx cos6 x sin x dx

2

1 u

6 ( du )

u

2

u 1

6 du

u

4 6

(u u ) du

Substitute u = cos x:

sin 3 x 1 cos 2 x

cos6 x dx cos6 x sin x dx

2

1 u

6 ( du )

u

2

u 1

6 du

u

4 6

(u u ) du

35.
TRY SUBSTITUTION Example 2

x

e dx

According to (ii) in Step 3 d, we substitute u = √x.

Then, x = u2, so dx = 2u du and e x dx 2 ueu du

The integrand is now a product of u and

the transcendental function eu.

So, it can be integrated by parts.

x

e dx

According to (ii) in Step 3 d, we substitute u = √x.

Then, x = u2, so dx = 2u du and e x dx 2 ueu du

The integrand is now a product of u and

the transcendental function eu.

So, it can be integrated by parts.

36.
RATIONAL FUNCTIONS Example 3

5

x 1

x3 3x 2 10 x dx

No algebraic simplification or substitution is obvious.

So, Steps 1 and 2 don’t apply here.

The integrand is a rational function. So, we apply

the procedure of Section 7.4, remembering that

the first step is to divide.

5

x 1

x3 3x 2 10 x dx

No algebraic simplification or substitution is obvious.

So, Steps 1 and 2 don’t apply here.

The integrand is a rational function. So, we apply

the procedure of Section 7.4, remembering that

the first step is to divide.

37.
TRY SUBSTITUTION Example 4

dx

x ln x

Here, Step 2 is all that is needed.

We substitute u = ln x, because its differential is

du = dx/x, which occurs in the integral.

dx

x ln x

Here, Step 2 is all that is needed.

We substitute u = ln x, because its differential is

du = dx/x, which occurs in the integral.

38.
MANIPULATE INTEGRAND Example 5

1 x

dx

1 x

Although the rationalizing substitution 1 x

u dx

works here [(ii) in Step 3 d], it leads to 1 x

a very complicated rational function.

An easier method is to do some algebraic manipulation

(either as Step 1 or as Step 4 c).

1 x

dx

1 x

Although the rationalizing substitution 1 x

u dx

works here [(ii) in Step 3 d], it leads to 1 x

a very complicated rational function.

An easier method is to do some algebraic manipulation

(either as Step 1 or as Step 4 c).

39.
MANIPULATE INTEGRAND Example 5

Multiplying numerator and denominator

by 1 x , we have:

1 x 1 x

dx dx

1 x 1 x2

1 x

dx 1 x dx

2 2

1 x

1 2

sin x 1 x C

Multiplying numerator and denominator

by 1 x , we have:

1 x 1 x

dx dx

1 x 1 x2

1 x

dx 1 x dx

2 2

1 x

1 2

sin x 1 x C

40.
CAN WE INTEGRATE ALL CONTINUOUS FUNCTIONS?

The question arises:

Will our strategy for integration enable us to

find the integral of every continuous function?

For example, can we use it to evaluate ∫ ex2dx?

The question arises:

Will our strategy for integration enable us to

find the integral of every continuous function?

For example, can we use it to evaluate ∫ ex2dx?

41.
CAN WE INTEGRATE ALL CONTINUOUS FUNCTIONS?

The answer is ‘No.’

At least, we cannot do it in terms

of the functions we are familiar with.

The answer is ‘No.’

At least, we cannot do it in terms

of the functions we are familiar with.

42.
ELEMENTARY FUNCTIONS

The functions we have been dealing

with in this book are called elementary

For instance, the function

x2 1

f ( x) 3 ln(cosh x) xesin 2 x

x 2x 1

is an elementary function.

The functions we have been dealing

with in this book are called elementary

For instance, the function

x2 1

f ( x) 3 ln(cosh x) xesin 2 x

x 2x 1

is an elementary function.

43.
ELEMENTARY FUNCTIONS

If f is an elementary function, then f’

is an elementary function.

However, ∫ f(x) dx need not be

an elementary function.

If f is an elementary function, then f’

is an elementary function.

However, ∫ f(x) dx need not be

an elementary function.

44.
ELEMENTARY FUNCTIONS

Consider f(x) = ex2.

Since f is continuous, its integral exists.

x

t2

If we define the function F by F ( x )

e

0

dt

then we know from Part 1 of the Fundamental2

Theorem of Calculus (FTC1) that F '( x) e x

Thus, f(x) = ex2 has an antiderivative F.

Consider f(x) = ex2.

Since f is continuous, its integral exists.

x

t2

If we define the function F by F ( x )

e

0

dt

then we know from Part 1 of the Fundamental2

Theorem of Calculus (FTC1) that F '( x) e x

Thus, f(x) = ex2 has an antiderivative F.

45.
ELEMENTARY FUNCTIONS

However, it has been proved that F is not

an elementary function.

This means that, however hard we try, we will never

succeed in evaluating ∫ ex2dx in terms of the functions

we know.

In Chapter 11, however, we will see how to express

∫ ex2dx as an infinite series.

However, it has been proved that F is not

an elementary function.

This means that, however hard we try, we will never

succeed in evaluating ∫ ex2dx in terms of the functions

we know.

In Chapter 11, however, we will see how to express

∫ ex2dx as an infinite series.

46.
ELEMENTARY FUNCTIONS

The same can be said of the following

x

e 2 x

x dx sin( x ) dx cos(e ) dx

3 1 sin x

x 1 dx ln x dx x dx

The same can be said of the following

x

e 2 x

x dx sin( x ) dx cos(e ) dx

3 1 sin x

x 1 dx ln x dx x dx

47.
ELEMENTARY FUNCTIONS

In fact, the majority of elementary

functions don’t have elementary

You may be assured, though, that the integrals

in the exercises are all elementary functions.

In fact, the majority of elementary

functions don’t have elementary

You may be assured, though, that the integrals

in the exercises are all elementary functions.