Chapter 5

New Words

Reduction formula 递推公式

Recursion formula 递推公式

Integration by parts 分部积分法

Integrate 积分 radical 根号

Perfect square 完全平方

Hyperbolic substitution 双曲替换

xvxu

xxuxvxxvxu

and functions abledifferenti two

any for d of in terms d

expressing of way a developssection This

.d findingthan

easier is d finding that hope We

xxvxu

xxuxv

This method is called integration by parts.

function rictrigonomet

inversean or ,sin,ln, form

thehas integrand when the wellespecially It works

xxxxex nnxn

uvuvvu

xxuxvxvxuxxvxu

d dor

1 d d

equation on the based is partsby n Integratio

We will first illustrate the method by an example.

Right after the example we will explain why (1) is

valid

1. Integration by parts

Solution

Question? ?dxxe x

is thisdo way toOne .d as d write

must wed d formula theuse To

vuxxe

uvuvvu

x

x

xx

x

x

ev

evex

v

xevv

xevxu

, of tiveantiderivaan is is, that ,d

d

,dd Since itself. findmust wecourse Of

.dd,

By integration by parts,

Cexexexexxe xxxxx dd

xxx xeCexe is of derivative thecheck thatmay You

Just as the chain rule is the basis for integration by

substitution, the formula for the derivative of a

product is the basis for integration by parts.

2. The proof of integration by parts

xxvxuxvxuxvxu

xvxuxvxuxvxu

d

uvuvvu

xxuxvxvxuxxvxu

d dor

d d

thatus tellsformula above theRewriting

3. Examples of integration by parts

Example 1 Find .cos xdxx

Solution (1) Let ,cos xu dvdxxdx 2

21

xdxxcos xdxx

xx

sin2

cos2

22

Obviously, if we choice impropriety is difficult to integral

vu ,

Solution (2) Let ,xu dvxdxdx sincos

xdxxcos xxd sin xdxxx sinsin

.cossin Cxxx

Remark:

(2) du should not be messier than u.

vuuv d original theneasier tha be should d 3

:met be

should conditions hree Usually t.d and labeling

theis partsby n integratio applying key to The

vu

messy. toobe

not should and gintegratinby found becan 1 v

Example 2 Find .2 dxex x

,2xu ,dvdedxe xx

dxex x2 dxxeex xx 22

.)(22 Cexeex xxx

（ Using again the integration by parts ） ,xu dvdxe x

Summary

If the integrand is the product of power functions and sine or cosine functions, power functions and exponential functions, we can let u as power function.

Solution

Example 3 Find .arctan xdxx

Solution: Let ,arctan xu dvx

dxdx 2

2

xdxxarctan )(arctan2

arctan2

22

xdx

xx

dxx

xx

x2

22

11

2arctan

2

dxx

xx

)1

11(

21

arctan2 2

2

.)arctan(21

arctan2

2

Cxxxx

Example 4 Find .ln3 xdxx

Solution: ,ln xu ,4

43 dv

xddxx

xdxx ln3 dxxxx 34

41

ln41

.161

ln41 44 Cxxx

Summary

If the integrand is the product of power function and

logarithm function, or power function and inverse

trigonometric function, we can let u as logarithm

function or inverse trigonometric function

In the following examples one integration by parts

appears at first to be useless, but two in succession

find the integral.

Example 5 Find .)sin(ln dxx

Solution:

dxx)sin(ln )][sin(ln)sin(ln xxdxx

dxx

xxxx1

)cos(ln)sin(ln

)][cos(ln)cos(ln)sin(ln xxdxxxx

dxxxxx )sin(ln)]cos(ln)[sin(ln

dxx)sin(ln .)]cos(ln)[sin(ln2

Cxxx

Example 6 Find .sin xdxe x

Solution: xdxe x sin xxdesin

)(sinsin xdexe xx

xdxexe xx cossin xx xdexe cossin

)coscos(sin xdexexe xxx

xdxexxe xx sin)cos(sin circulation

xdxe x sin .)cos(sin2

Cxxe x

Example 7 Find .

1

arctan2

dxx

xx

Solution: ,1

12

2

x

xx

dx

x

xx21

arctan 21arctan xxd

)(arctan1arctan1 22 xdxxx

dxx

xxx 222

11

1arctan1

dxx

xx

2

2

1

1arctan1 Let tx tan

dxx 21

1

tdtt

2

2sec

tan1

1 tdtsec

Ctt )tanln(sec Cxx )1ln( 2

dx

x

xx21

arctan

xx arctan1 2 .)1ln( 2 Cxx

Reduction formulas

We can get some reduction formulas or recursion

formulas by an integration by parts.

Solution:

xxI nn dsin FindExample 8

xxxn

xxxxI

n

nnn

dcossin1

sincoscosdsin

22

11

xxxnxx nn dsin1sin1sincos 221

nnn InInxx 11sincos 2

1

xxn

In

nI

xxInnI

nnn

nnn

12

12

sincos11

,sincos1 is,That

2cossin1

dcos1

dcos

:follows assimilarly

obtained is dcos of formulareduction The

12

nxxn

xxn

nxx

xx

nnn

n

Solution:

x

axI

nn d1

Find22

Example 9

Ca

x

ax

axIn

arctan

1d

1 ,1When

221

xax

xn

ax

x

xax

I

n

nn

nn

d12

d1

yields partsby n integratio ,1When

22

2

122

1221

xax

an

xax

nax

x

xax

aaxn

ax

x

n

nn

nn

d1

12

d1

12

d12

22

2

122122

22

222

122

nnnn IanIn

ax

xI 2

11221 1212

is,That

1,32

12

111222

nIn

ax

x

anI nnn

Solution:

,d)()()(dd)( xxfxxfxfxxxfx

xxfx

xfe x

d)( find

, of tiveantiderivaan is that Suppose2

Example 10

Ceexdxxfxxfdxxfx

edxxfxeexf

xx

xxx

22

222

22)()()(

,C )( and 2

,assumptiongiven Under the