chapter 5. new words reduction formula 递推公式 recursion formula 递推公式 integration by...
TRANSCRIPT
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Chapter 5
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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
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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.
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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
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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,
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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
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xxvxuxvxuxvxu
xvxuxvxuxvxu
d
uvuvvu
xxuxvxvxuxxvxu
d dor
d d
thatus tellsformula above theRewriting
3. Examples of integration by parts
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Ceexdxxfxxfdxxfx
edxxfxeexf
xx
xxx
22
222
22)()()(
,C )( and 2
,assumptiongiven Under the