27

Chapter 8: TECHNIQUES OF INTEGRATION

BASIC INTEGRATION FORMULAS TABLE:

Differentiation Formula Integration Formula

1

1)(

1

ncn

xdxxx

nn

)(')()]([ 1 xuxnuxudx

d nn 1

)(')(1

n

uduudxxuxu

nnn

+C 1n

)(')(cos))((sin xuxuxudx

d Cxuududxxuxu )(sincos)(')(cos

)(')(sin))((cos xuxuxudx

d Cxuududxxuxu )(cossin)(')(sin

)(')(sec))((tan 2 xuxuxudx

d Cxuduudxxuxu )(tansec)(')(sec 22

)(')(csc))((cot 2 xuxuxudx

d Cxuduudxxuxu )(cotcsc)(')(csc 22

)(')(tan)(sec))((sec xuxuxuxudx

d Cxuduuudxxuxuxu )(sectansec)(')(tan)(sec

)(')(cot)(csc))((csc xuxuxuxudx

d Cxuduuudxxuxuxu )(csccotcsc)(')(cot)(csc

)(')( )()( xueedx

d xuxu cedxxue xuxu )()( )('.

)(

)('))((ln

xu

xuxu

dx

d cxudu

xudx

xu

xu)(ln

)(

1

)(

)('

0ln)(')( )()( aaxuaadx

d xuxu ca

adxxua xuxu )()(

ln

1)('.

)(1

)('))((tan

2

1

xu

xuxu

dx

d

ca

u

aua

dudx

ua

u 1

2222tan

1'

)(1

)('))((sin

2

1

xu

xuxu

dx

d

ca

u

ua

dudx

ua

u

1

2222sin

'

1)(

)('))((sec

2

1

xuu

xuxu

dx

d c

a

u

aauu

dudx

auu

u

1

2222sec

1'

28

Sec 8.1: Method of INTEGRATION BY PARTS

From the product rule, we can obtain the following formula, which is very useful in integration:

Start with vdx

du

dx

dvuxvxu

dx

d))'()(( .

Integrate both sides to get vduudvuv

Rearrange to obtain the integration by parts formula

To integrate by parts, strategically choose u, dv and then apply the formula.

Key Concept

Choose u, dv in such a way that:

o u is easy to differentiate.

o dv is easy to integrate.

o

Sometimes it is necessary to integrate by parts more than once

ILATE are supposed to suggest the order in which you are to choose the “u:

for choosing which of two functions is to be u and which is to be dv is to choose u by whichever

function comes first in the following list:

I: inverse trigonometric functions: arctan x , arcsec x, etc.

L: the logarithmic function: ln x

A: algebraic functions: x2,3x50, etc.

T: trigonometric functions: sin x, tan x, etc.

E: exponential functions: ex, 13x, etc.

Integration by parts ``works'' on definite integrals as

well:

29

Exercise: # 8 Evaluate dxxex3

Exercise # 2 Evaluate dcos

_________________________________________________________________________________

Exercise: # 6 e

xdxx1

3ln # 14 dxxx 2sec4 2

30

Examples: Repeated integration by parts

# 4 Evaluate dxxx sin2 # 10 Evaluate dxexx

x22 )12(

# 21 de sin

31

Example: We can also sometimes use integration by parts when we want to integrate a functions such

as the inverse functions xx ln&.....,sin 1

that cannot be split into the product of two things. The trick we use is to take dv = 1.

Example: : Evaluate dxxln # 12 Evaluate

dyy1sin

32

Tabular Integration by parts:

it only works when a polynomial function is chosen to be the u.

This technique is best illustrated with an example. Let us consider a function that would be incredibly difficult

to do with our usual technique for integration by parts.

Example: Example :

Determine dxexx2 Determine dxxx sin2

?

Using tabular integration by parts

Using tabular integration by parts

Example :

Evaluate .

Solution:

Remember, this technique only works if a polynomial function is going to be selected as u. Just because an

integral contains a polynomial does not mean that the tabular method can be applied.

For example, one could not apply this method to the integral in Example 2. In that instance, LIPET told us to

select ln(x) as u, not x3. (In fact, only one use of integration by parts was required to solve that integral.)

Signs: Differentiate: Integrate:

+ x2 2x ex

- 2x ex ex

+ 2 2 ex

- 0 e ex x

Signs Differentiate:: Integrate:

+ x2

x xx2 sin(x)

- 2x - xcos cos(x)

+ 22 2 -sin(x)

- 0 cos xcos (x)

33

Sec 8.2: Trigonometric integrals

These are integrals of the following form:

dxxmsin , dxxn

cos and dxxx nm cossin

We have TWO cases,

Case I: if n OR m is an odd positive integer:

we will be using the identity ,cossin 122 xx

xxandxxwhere2222 11 sincos,cossin

If n is odd: Save one xcos out and use the identity xx22

sin1cos to change the

remaining function to xsin , then use the substitution xu sin and xdxdu cos .

Example: dxxx43

sincos

34

If m is odd: Save one xsin out and use the identity xx22

cos1sin to change the

remaining function to xcos , then use the substitution xu cos and xdxdu sin .

Example: dxxx34

sincos

35

Case II: m and n are both even positive integers

We will be using the identities:

# 14 2/

0

2sin

dxx Exercise # 17

0

4sin8 dxx

36

Integration of Powers of Sec and tan , cot and csc

These are integrals of the following form:

dxxnsec , dxxm

tan , dxxx mn tansec and dxxx mn

cotcsc .

For integrals of this form, you will either integrate by substitution, letting tanu x x or

secu x x , or you will use integration by parts(Reduction formula). Which path you take

depends on whether m or n is even or odd…

m n u(x)

Even tanu x x ; see Case 1

Odd secu x x ; see Case 2

Even Even tanu x x ; see Case 3

Even Odd Use integration by parts(reduction formulas;)

see Case 4

So we have three cases, in all the three cases , we will be using the identity

1costan,sec1tan2222 xxwherexx

Case 1: When n is even positive integer , Save x2

sec out and use the

identity 1tansec22 xx to change the remaining function to xtan , then use the

substitution xu tan and xdxdu2

sec .

Example: Exercise # 38 xdxx24 tansec

37

Case 2: When m is odd positive integer ,Save one xx tansec out and use the

identity 1sectan22 xx to change the remaining function to xsec , then use the

substitution xu sec and xdxxdu tansec .

Example: Exercise # 36 dxxx33

tansec

38

Case 3: dxxm tan where m is even

Example: dxx4

tan

Case 4 : dxxnsec ,where n is odd

dxx3

sec: Example

39

Sec 8.3: Trigonometric Substitution:

Integrals involving integrands with 2 2a x or 2 2x a .

So there are cases as follow

Case 1: Integrals involving 2 2a x

1) Make the substitution x = asin(), and dx = acos()d.

2) Simplify the radical using algebra and the Pythagorean identity:

22 2 2 2 2sin 1 sin cos cosa a a a a

3) Integrate.

4) To back-substitute, use the following trick from trigonometry.

Since x = asin() sin() = x / a, which means we can draw the diagram below

to evaluate any trigonometric function of :

Example:

Evaluate

dxx

21

1

Let ddxthenx cossin

ccdd

d

ddxx

1

2

2

22

sin1coscos

1

cossin1

1

cossin1

1

1

1

40

Case 2: Integrals involving 2 2a x

1) Make the substitution x = atan(), and dx = asec2()d.

2) Simplify the radical using algebra and the Pythagorean identity:

22 2 2 2 2tan 1 tan sec seca a a a a

3) Integrate.

4) To back-substitute, use the following trick from trigonometry.

Since x = atan() tan() = x / a, which means we can draw the

diagram below to evaluate any trigonometric function of :

Example: .

Evaluate

dxx

21

1

Let ddxthenx2sectan

1

2

2

2

22

tan

1

secsec

1

sectan1

1

1

1

cd

ddxx

41

Case 3: Integrals involving 2 2x a

1) Make the substitution x = asec(), and dx = asec()tan()d.

2) Simplify the radical using algebra and the Pythagorean identity :

2 2 2 2 2 2sec sec 1 tan tana a a a a .

3) Integrate.

4) To back-substitute, use the following trick from trigonometry.

Since x = asec() sec() = x / a, which means we can draw the

diagram below to evaluate any trigonometric function of :

525

3

2

ydyy

y# 12

42

Exercise # 2

dxx

291

3 # 4

2

02

28 x

dx

# 8 #

dxx

x

12

3

31

dxt291

43

Exercise # 24

1

02/32 )4( x

dx

use an appropriate substitution and the a trigonometric substitution evaluate the integral

4ln

0 29

t

t

e

dte35)

)3/4ln(

)4/3ln(2/32

)1(t

t

e

dte36 #

44

Sec 8.4:Integration of Rational functions by Partial fractions

into its )(

)(

xQ

xPf by decomposing

)(

)(

xQ

xPs of the form Our goal in this section is to compute integral

partial fractions ,

where )(xP and )(xQ are polynomials and

Degree of )(xP is < Degree of )(xQ

For Example :

dxxx

x

32

352

In such a case we express the algebraic fraction in terms of its Partial fractions , as follows

3

3

1

2

32

352

xxxx

x

so dxx

dxx

dxxx

x

3

3

1

2

32

352

cxx )3ln(3)1ln(2

Now how do we find Partial Fractions of )(

)(

xQ

xP

Example:

The Rules of Partial fractions are as follows

(1) The numerator )(xP must be of lower degree than that of the denominator )(xQ . If it is not,

Use long division to ensure that the degree of )(xP less than the degree of )(xQ .

(2) Factorize the denominator )(xQ as far as possible. This is important since the factors obtained

determine the shape of the partial fractions.

(3) This fraction can be broken down into partial fractions, that is dependent upon the factors of the

denominator , there are four cases:

(4) linear factors dcx

B

bax

A

dcxbax

xP

))((

)(.

(5) Repeated Linear Factors 22 )()(

)(

bax

B

bax

A

bax

xP

.

(6) More Repeated linear Factors 323 )()()(

)(

bax

C

bax

B

bax

A

bax

xP

.

(7) Quadratic factor edx

C

cbxax

BAx

edxcbxax

xP

22 ))((

)(

2223 22243

4

)())(( xxxxx

x

45

Next How do we find A,B,C ……

Consider the example above

3132

352

x

B

x

A

xx

x

Multiply both sides by the denominator )3)(1( xx .

)1()3(35 xBxAx

Now we can solve for A and B using one of the following methods:

Substitution Or Equate coefficients )1()3(35 xBxAx BBxAAxx 335

put 3x )2(33

)1(5

BA

BA

34315 BB solve eq's (1) & (2) 3255

284

BAB

AA

Put 1x

248 AA

So 3

3

1

2

32

352

xxxx

x

__________________________________________________________________________________

# 10 xx

dx

22 # 16

dx

xx

x

82

33

46

#22

3

13

2 43dt

tt

tt #24

dx

x

xx22

2

)14(

288

# 28 dxxx

4

1 #30

dx

xx

xx

43 24

2

47

# 34

dxx

x

12

4

#40

dt

e

eeet

ttt

1

22

24

(DEGREE OF NUMERATOR > DEGREE OF DENIMINATOR)

48

Sec 8.7: Improper Integrals

In this section we need to take a look at a couple of different kinds of integrals. Both of these are examples of

integrals that are called Improper Integrals.

First Type: integrals in which one or both of the limits of integration are infinity so they are of the

forms

(1)

a

dxxf )(

(2)

b

dxxf )(

(3)

dxxf )(

and they can be evaluated using the following definitions :

(1) If )(xf is continuous on ),[ a , then

R

aa Rdxxfdxxf )()( lim

(2) If )(xf is continuous on ],( b , then

a

R

a

Rdxxfdxxf )()( lim

(3) If )(xf is continuous for all ),( x and if c is any real number, then

which can be evaluated using (1) and (2).Note: If either of the integrals in (3)

diverge, then

dxxf )( diverge.

Example 1: Evaluate the

following integral.

12

1dx

x

Solution

c

c

dxxfdxxfdxxf )()()(

So, this is how we will deal with

these kinds of

integrals in general. We will

replace the

infinity with a

variable (Like b),

do the integral and then take the

limit of the result

as b goes to

infinity.

notice that the

area under a

curve on an

infinite interval was not infinity

as we might have

suspected it to

be. In fact, it was a surprisingly

small number.

Of course this won’t always be

the case, but it is

important enough to point out that

not all areas on

an infinite

interval will yield infinite areas.

Let’s now get

some definitions

out of the way. We will call these

integrals

convergent if the associated limit

exists and is a

finite number (i.e. it’s not plus or

minus infinity)

and divergent if

the associated limits either

doesn’t exist or is

(plus or minus)

infinity.

Example 1: Evaluate the

following integral.

1

1dx

x

Solution

49

Example: Determine if the following integral is Example: Determine if the following integral is

convergent or divergent. If it is convergent or divergent. If it is

convergent find its value. convergent find its value.

0

4dxxe

x

dx

x2

1

1

# 14

2/32 )4(x

xdx # 24

dxxe

x2

2

50

# 20

02

1

1

tan16dx

x

x # 19

012 )tan1)(1( vv

dv

# 12

22 1

2

t

dt # 2

1001.1

x

dx

.

51

Second Type: integrals that have discontinuous integrands f(x)

The process here is basically the same with first kind. Here are the general cases that we’ll look at for these

integrals.

(1) If )(xf is continuous on ),[ ba and discontinuous at b , then

R

a

b

a bR

dxxfdxxf )()( lim

(2) If )(xf is continuous on ],( ba and discontinuous at a , then

b

R

b

a aR

dxxfdxxf )()( lim

(3) If )(xf is discontinuous at a number ],[ bac , then

b

c

c

a

b

a

dxxfdxxfdxxf )()()(

Note: If either of the integrals in (3) diverge, then b

a

dxxf )( diverges

(4) If )(xf is discontinuous at a and b , then

b

c

c

a

b

a

dxxfdxxfdxxf )()()(

Note: If either of the integrals in (4) diverge, then b

a

dxxf )( diverges.

Note that the limits in these cases really do need to be right or left handed limits. Since we will be working

inside the interval of integration we will need to make sure that we stay inside that interval. This means that

we’ll used one-sided limits to make sure we stay inside the interval.

b

RcR

R

acR

dxxfdxxf )()( limlim

R

cbR

c

RaR

dxxfdxxf )()( limlim

52

Example: Determine if the Example: 2/

0

tan

xdx

following integral is

convergent or divergent. If it is

convergent find its value.

5

1 5

2dx

x

# 16

2

0 24

1ds

s

s # 26

1

0

)ln( dxx

53

Before leaving this section lets note that we can also have integrals that involve both of

these cases First kind and Second kind. Consider the following integral.

Example: Determine if the following integral is convergent or divergent. If it is

convergent find its value.

3

1

x

Solution This is an integral over an infinite interval that also contains a discontinuous integrand. To do this integral

we’ll need to split it the given integrals as follows :

In order for the integral in the example to be convergent we will need ALL of these to be convergent. If

one is divergent then the whole integral will also be divergent.

So, the first integral is divergent and so the whole integral is divergent.