Chapter 6

6.4 Integration of substitution and int6.4 Integration of substitution and integration by parts of the definite integregration by parts of the definite integralal

As we all known, the integration of substitution ( or

“change of variables”) and integration by parts are very

important tools to evaluate the indefinite integrals. In

this section we extend these methods to the definite

integrals.

1. Integration of substitution1. Integration of substitution

At first, the substitution technique extends to definite

integrals

Theorem 1Theorem 1 (Substitution in a definite integral)

b

abatt

t

tx

baxf

, and ,],[],,[ when )3(

0 and

,on derivative continuous a has 2

;,on continuous is 1

:satisfied are conditions following theIf

tttfxxf

txb

ad)()]([d)(Then

Substitution formula of definite integrals

][tLet ),()()(then

, of tiveantiderivaan is that Suppose

tFaFbFdxxf

xfxFb

a),()]([)()(

d

d

d

d)( ttftxf

t

x

x

Ft

b

axxfaFbFFF

tttf

d)]([)]([

)()(d)()]([

ttf ][ of tiveantiderivaan is t

ProofProof

NoticeNotice

One important is the necessary change in the limits of

integration. The following examples will apply this

technique.

Example 1Example 1 .dsincos Evaluate 2

0

5

xxx

SolutionSolution

,dsind then ,cosLet xxtxt

Thus 1. to0 from

goes cos ,2

to0 from goes as e,Furthermor xtx

.6

1

6ddsincos

1

0

60

1

5

0

52 t

ttxxx

Example 2Example 2

SolutionSolution

..dsinsin Evaluate0

53

xxx

02

3

0

53

2

353

dsincosdsinsin

sincossinsin)(

xxxxxx

xxxxxf

2

2

32

02

3

dsincosdsincos xxxxxx

Additivity over intervals

2

2

32

02

3

sindsinsindsin xxxx

.5

4sin

5

2sin

5

2

2

2

2

5

0

2

5

xx

Change of variables

Newton-Leibniz formula

Example 3Example 3

SolutionSolution

.)ln1(ln

d Find

4

3

e

e xxx

xI

43

)lnarcsin(2e

ex .6

4

3

4

3

2)ln(1

lnd2

)ln1(ln

lnd e

e

e

e x

x

xx

xI

Example 4Example 4

SolutionSolution

a

axxax

I0 22

)0(,d1

Find

,00,2

and ,dcosd then ,sinLet

txtax

ttaxtax

2

0

2

0

2

0 22

dcossin

sincos1

2

1d

cossin

cos

d)sin1(sin

cos

ttt

ttt

tt

t

ttata

taI

.4

cossinln2

1

22

1 20

tt

Example 5Example 5 4

1 if ,d2 Find xxf

SolutionSolution

01,cos1

10,

2

xx

xxexf

x

thus,,24

,11 and ,dd then ,2Let

tx

txtxtx

2

0

0

1

2

1

4

1 d d d d2 ttfttfttfxxf

2

0

0

1 de d

cos1

1 2

tttt

-t

2

1

2

1

2

1tan

2

1

2tan 4

2

0

0

1

2

eet t

2. Important simplification formulas2. Important simplification formulas

Theorem 2Theorem 2

,d)(2d)(

then function,even an is If 1

thatshow

,,on continuous is that Suppose

0

aa

axxfxxf

xf

aaxf

,d)(d)(d)(0

0

a

a

a

axxfxxfxxf

0d)( then function, oddan is If 2 a

axxfxf

SolutionSolution

then,let ,d)( evaluate order toIn 0

txxxfa

,d)(

d)(d)(d)(

0

0

00

a

a

aa

xxf

ttfttfxxf

aaa

axxfxxfxxf

00d)(d)( d)(

a

xxfxf0

d)(

,d)(2d)(

,2then

, then function,even an is If 1

0

aa

axxfxxf

xfxfxf

xfxfxf

0d)(

,0then

, then function, oddan is If 2

a

axxf

xfxf

xfxfxf

NoticeNotice

The properties of definite integrals for odd and even

functions provide a easy way to evaluate their

definite integrals. The following examples will apply

the property.

Example 6Example 6 .d11

cos2 Evaluate

1

1 2

2

x

x

xxx

SolutionSolution

1

1 2

2

d11

cos2x

x

xxx

1

1 2

2

d11

2x

x

x

even function

1

1 2d

11

cosx

x

xx

1

0 2

2

d11

4 xx

x

1

0 2

22

d)1(1

)11(4 x

x

xx

.4

odd function

1

0

21

0

2 d144d)11(4 xxxx

Area of unit circle

Example 7Example 7 .dsin1 Find 2

2

3202

xxx

2

2

3202 dsin1

xxx

SolutionSolution odd function

0

Example 8Example 8

? then ,dcossin

,dcossin

,dcos1

sin that Suppose

2

2

432

2

2

43

2

2

42

xxxxP

xxxN

xxx

xM

NMPDPMNC

NPMBMPNA

;

; ;

SolutionSolution

0dcos1

sin

2,

2on function oddan is cos

1

sin

2

2

42

42

xxx

xM

xx

xxf

0dcos2dcossin and 2

0

42

2

43

xxxxxN

0dcos2dcossin 2

0

42

2

432

xxxxxxP

Thus, we choice ( D )

Theorem 3Theorem 3

TTx

xttfttf

Txf

0dd that prove

, period offunction periodic a is that Suppose

ProofProof

Tx

T

T

x

Tx

xttfttfttfttf dddd

0

0

T

xxT

dttf

dssfdttfdttf

0

000

x

x

TdssTfdttfdttf

0

0

0 f (x+T)=f (x)

The integral properties of periodic functions provide

also a easy way to evaluate some definite integrals.

Next example will apply the property.

Example 9Example 9

k

kxxxI dsinsin Evaluate 53

SolutionSolution

2

2

3

2

0

2

3

0

2

3

dcossindcossin

dcossin

xxxxxx

xxxI

period offunction

period a isit ,cossincossin

and ,cossinsinsin

2

3

2

3

2

353

xxxx

xxxx

5

4sindsinsindsin

2

2

3

2

0

2

3

xxxx

ProofProof（ 1） Let tx 2

,dtdx

0x ,2 t

2x ,0 t

0 2

00

00

dcos1

sin find and

d)(sin2

d)(sin 2

;d)(cosd)(sin 1

thatprove [0,1],on continous is If

22

xx

xx

xxfxxxf

xxfxxf

xf

Example 10Example 10

2

0d)(sin

xxf

,0 and ,0

,ddLet 2

txtx

txtx

0

2

d2

sin

ttf

2

0d)(cos

ttf ;d)(cos2

0

xxf

0

0d)][sin()(d)(sin

ttftxxxf

,d)(sin)(0 ttft

00

d)(sind)(sin tttfttf

,d)(sind)(sin00

xxxfxxf

.d)(sin2

d)(sin00

xxfxxxf

00 2

0 20 2

)arctan(cos2

)(cosdcos1

1

2

dcos1

sin

2d

cos1

sin

formula, above theapplyingBy

xxx

xx

xx

x

xx

.4

)44

(2

2

3 Integration by parts3 Integration by parts

Theorem 4Theorem 4

Next, the integration by parts extends to definite

integrals

b

a

b

a

b

auvuvvu

ba

xvxu

dd

then,,on derivative

and continuous are and that Suppose Integration by parts for definite integrals

,d)( and ,b

a

b

auvxuvvuvuuv

,dd

equation,

above of sidesboth integral definite theTaking

b

a

b

a

ba xvuxvuuv

.dd b

a

b

a

b

auvuvvu

ProofProof

This completes the proof

SolutionSolution

Example 11Example 11 .darcsin Evaluate 21

0 xx

,,1

dd

then,dd ,arcsin Setting

2xv

x

xu

xvxu

)1(d1

1

2

1

62

11

darcsindarcsin

2

0 2

0 200

21

21

21

21

xx

x

xxxxxx

.12

3

121

1221

02

x

Example 12Example 12 ..2cos1

d Evaluate 4

0

x

xx

SolutionSolution

,cos22cos1 2 xx

xx

x

xx

x

xxtand

2cos2

d

2cos1

d 444

00 20

xxxx dtan2

1tan

2

1 44

00

.4

2ln

8secln

2

1

840

x

Example 13Example 13 ...d)2(

)1ln( Evaluate

1

0 2

xx

x

SolutionSolution

1

0

1

0 2 2

1d)1ln(d

)2(

)1ln(

xxx

x

x

1

0

1

0

)1ln(d2

1

2

)1ln(x

xx

x

xx

21

11x

xxd

1

1

2

1

3

2ln 1

0

.3ln2ln3

5)2ln()1ln(

3

2ln 10 xx

Example 14Example 14

constant anot 0;

constant; negative a constant; positive a

? is then ,dsineLet 2 sin

DC

BA

xFttxFx

x

t

SolutionSolution

A

2

0

sin2 sin

sin

dsine dsine

,2 periodith function w

periodic a is sine integrand theSince

ttttxF

xxf

tx

x

t

x

2

0

sin2

0

sin

2

0

sin

edcosecos

cosde

tt

t

tt

t

0dcose2

0

2sin

ttt

Thus, we choose (A)

.d)( find ,dsin

)( that Suppose1

01

2

xxxftt

txf

x

Example 15Example 15

SolutionSolution

directly, find way tono tive,antideriva

elementaryan no has sin

function theSince

xft

t

1

0

21

0

21

02

1

0

21

0

d)(2

1)1(

2

1)(d

2

1)(

2

1

)(d)(2

1d)(But

xxfxfxfxxfx

xxfxxxf

,sin2

2sin

)(

,0dsin

)1( ,dsin

)( and

2

2

2

1

11

2

x

xx

x

xxf

tt

tft

t

txf

x

1

0

21

0d)(

2

1)1(

2

1d)( xxfxfxxxf

).11(cos2

1cos

2

1

dsin2

1dsin2

2

1

1

02

1

0

221

0

2

x

xxxxx

number oddan is 1 ,3

2

5

4

2

31

numbereven an is ,22

1

4

3

2

31

dcosdsin that Prove 22

00

nn

n

n

n

nn

n

n

n

xxxxI nnn

Example 16Example 16

SolutionSolution

,cos ,dcossin)1(d

then,dsind ,sin Setting2

1

xvxxxnu

xxvxun

n

xxxnxxI nnn dcossin)1(cossin 22

0

220

1

x2sin1 0

nn

nnn

InIn

xxnxxnI

)1()1(

dsin)1(dsin)1(

2

00

2 22

,2

3 and ,

1 Thus 422

nnnn I

n

nII

n

nI

nI of formulareduction The

,2,1 where

,3

2

5

4

7

6

12

22

12

2

,2

1

4

3

6

5

22

32

2

12

112

02

m

Im

m

m

mI

Im

m

m

mI

m

m

thus,,1dsin,2

d and 22

0100

xxIxI

.3

2

5

4

7

6

12

22

12

2

,22

1

4

3

6

5

22

32

2

12

12

2

m

m

m

mI

m

m

m

mI

m

m