Chapter 4: Methods of integrals

Definition

Section 4.1: Indefinite Integrals

The process of finding the indefinite integral of f(x) is called integration of f(x) or integrating f(x). If we need to be specific about the integration variable we will say that we are integrating f(x) with respect to x.

Properties of the Indefinite Integral:

need. weas functionsmany as to

extended becan rule Thisintegrals. individual theof difference

or sum theis functions of differenceor sum a of integral the

s,other wordIn 2 .)()()()( dxxgdxxfdxxgxf

integrals. indefinite ofout constants tivemultiplicafactor can

weSo, number.any iswhere1 ,)()(. kdxxfkdxxfk

.)()()()( dxxgdxxfdxxgxf

.)(

)(

)(

)(

dxxg

dxxfdx

xg

xf

!!!!! Warning !!!!!

.1,1

1

ncn

xdxx

nn

.constants areand, kcckxkdx

.cossin cxxdx .sincos cxxdx

.tansec2 cxxdx .sectansec cxxdxx

The first integral that we will look at is the integral of a power of x.

The general rule when integrating a power of x we add one onto the exponent and then divide by the new exponent. It is clear that we will need to avoid n = -1 in this formula. If we allow in this formula we will end up with division by zero.

Next is one of the easier integrals but always seems to cause problems for students.

Let us now take a look at the trigonometric functions:

Example:

.cedxe xx .||ln1 1 cxdxxdxx

.4105)( 63 dttta

.)( 88 dxxxb

.424

54

510

45

4105

5454

63

cttt

cttt

dtdttdtt

.7979

797988 c

xxc

xxdxxdxx

.1 cydy .)( dyc

Now, let us take care of exponential and logarithm functions.

.6

773)(

54 3 dx

xxxd dx

xdx

xdxx

1

6

7173

54 3

dxxdxxdxx 2

154

13

6

773

cxxx

1216

7

157

143

3

12

1151

4

3

cxxx

3

7

4

7

7

12 2

144

7

.4)( 23 dwwwwe

.1524

)(3

2610

dxx

xxxf

.44 3233 dwwwwww

dxxdxxdxx 137 1524

.44 3

7

3

13

dwwwww

.

310

34

442

43

10

3

442

cwwww

.10

33

42

3

10

3

442 c

ww

ww

cxxx

||ln154

28

448

cxxx

||ln1522

48

Integration by substitution:

dyewdwdtt

dxx ycos13

4

dyeydwww

wyy2418lncos

11

dttt

tdxxx

34

34 32

)2(

125618

After the last section we now know how to do the following integrals

However, we can’t do the following integrals:

Let us start with the first one

dxxx 4 32 5618

In this case let us notice

duudxxxdxxx 4

124

134 32 18565618

56 3 xu

and we compute the differential dxxdxudu 218'

Now, we go back to our integral and notice that we can eliminate every x that exists in the integral and write the integral completely in terms of u using both the definition of u and its differential

Evaluating the integral gives,

c

xc

uduudxxx

5

564

5

45618

4

534

5

4

14 32

Example:

Solution:

dwww

wa lncos

11)(

dxxxc432 103)(

dyeyb yy24183)(

dx

x

xd

241)(

(a) In this case let us take wwu lnand we compute the differential

dww

dwudu

11'

Now, we go back to our integral and notice that we can eliminate every w that exists in the integral and write the integral completely in terms of u using both the definition of u and its differential

cwwcuduu

dww

wwdwwww

lnsin)sin(cos

11lncoslncos

11

dyeyb yy24183)( In this case let us take yyu 24

and we compute the differential dyydyudu 18'

Thus we get

ceceduedyyedyey yyuuyyyy 222 444 333183183

In this case let us take3103 xu

So, dxxdxudu 230'

Thus we get

dxxxdxxx )30(103

30

1103 243432

dxxxc432 103)(

c

xcuduu

150

103

150

1

30

153

54

dx

x

xd

241)(

In this case let us take 241 xu

So, dxxdxudu 8'

Thus,

dxxxxdxxdx

x

x8)41(

8

1)41(

412

122

12

2

cu

duu

218

1

8

1 2

1

2

1

cx 2

12 )41(

4

1

Integration by parts:

Example 1:

Solution: dxxex Find

dxex x2 FindExample 2: Solution:

.,., Then Let xx evdxdudxedvxu

c. xxxxx exedxexedxxe

.,2., Then Let 2 xx evxdxdudxedvxu

dxxeexdxex xxx 222

c22 xxx exeex

C222 xxx exeex

Inserting this integral in the first one yields

Bringing the last term to the left hand side and dividing by 2 gives

Example 3:

Solution:

xdxex cos Find

.sin,.cos, Then Let xvdxeduxdxdveu xx

.sinsincos Thus, xdxexexdxe xxx

So,Then Let

. integral theagain partsby integrate Now

.cos,.sin,

sin

xvdxeduxdxdveu

xdxe

xx

x

.coscos)cos(cossin xdxexedxxexexdxe xxxxx

.coscossincos xdxexexexdxe xxxx

xexexdxe xxx cossincos2 .cossin

2

1cos cxexexdxe xxx

Example 4:

Solution:

.5

,1

.,ln 5

4 Then Letx

vdxx

dudxxdvxu

.1

5ln

5ln Thus,

554 dx

x

xx

xxdxx

.ln Find 4 xdxx

dxxxx 4

5

5

1ln

5

c55

1ln

5

55

x

xx

c.25

ln5

55

x

xx

Example 5:

Solution:.,

1..1,ln Then Let xvdx

xdudxdvxu

.1

lnln Thus, dxx

xxxxdx

.ln Find xdx

dxxx .1ln

cln xxx

Integration by partial fractions:

Main Rules:

Example 1:

First we write

Thus

Integrating, we get

Distinct roots:

)2)(1( Integrate

xx

xdx

Solution:

21)2)(1(

x

B

x

A

xx

x

getting ,)2)(1(by equation hismultiply t Now x-x-

)1()2( xBxAx

;1 so ,)21(1get we,1 substitute weIf -AAx

.2 so ,)12(2get we,2 letting Now BBx

2

2

1

1

)2)(1(

xxxx

x

.|2|ln2|1|ln)2(

2)1()2)(1(

cxxx

dx

x

dx

xx

xdx

Integrating, we obtain

Therefore

Example 2:

)3)(1(

)3( Integrate

2

2

xx

dxx

Solution:

)3)(1)(1( getting ,by equation hismultiply t Now xxx

;4/1 so ,)4)(2(3-1get we,1 Substitute -AAx

and writeFirst we ),1)(1()1( 2 xxx

)3()1()1()3)(1)(1(

)3(

)3)(1(

)3( 2

2

2

x

D

x

B

x

A

xxx

x

xx

x

)1)(1()3)(1()3)(1()3( 2 xxDxxBxxAx

;2/1 so ,)2)(2(3-1get we,1 Substitute BBx

;4/38/6 so ,)2)(4(3-9get we,3 Substitute CCx

)3(

1

4

3

)1(

1

2

1

)1(

1

4

1

)3)(1(

)3(2

2

xxxxx

x

.|3|ln4

3|1|ln

2

1|1|ln

4

1

)3)(1(

)3(2

2

cxxxxx

dxx

which we can integrate term by term

Thus

Therefore

Multiple roots:Example 3:

First we write

)1(

)1( Integrate

2

2

xx

dxx

Solution:

1)1(

)1(22

2

x

D

x

B

x

A

xx

x

1

211

)1(

)1(22

2

xxxxx

x

22 )1()1(1 DxxBxAxx

.1 so ,)1(1get we,0 ngSubstituti BBx

; of eother valuany take we find To .2get we,1 ngSubstituti xADx

.1soand,22-

so ,2)2)(1()2)(1(2get we,1 Substitute

AA

Ax

.|1|ln21

||ln)1(

)1(2

2

cxx

xxx

dxx

Example 4:

Solution:

First we write

Multiply by we get

Thus

Integrating, we obtain

8

3,So.

8

9

8

85121

8

5

2

33

,8

5

2

33331get we0 Substitute

AA

ADBA x

3)1(1)1)(3(

)1(22

2

x

D

x

B

x

A

xx

x

22 )1()3()1)(3(1 xDxBxxAx

.8

5,)4(19get we,3 Substitute 2 DsoD x

.2

1),4(2get we,1 Substitute BsoB x

3

1

8

5

)1(

1

2

1

1

1

8

3

)1)(3(

)1(22

2

xxxxx

x

.|3|ln8

5

)1(2

1|1|ln

8

3

)1)(3(

)1(2

2

cxx

xxx

dxx

Solution:

Example 5:

No roots:

Main Rule:

Thus

cx

x

dx

x

dx

)

3(tan

3

1

391

222

ca

x

aax

dx

)(tan

1 122

9 Evaluate

2x

dx

Solution:

Example 6:

1)2(45)44(54 222 xxxxx

54 Evaluate

2 xx

dx

Here we can’t find real factors, because the roots are complex. But we can complete the square:

.)2(tan1)2(54

122

cxx

dx

xx

dx