大一上微積分林國建課件-第五章5.4換元積分法

萬能替換對于由三角函數(shù)sin

x,cos

x,tan

x,cot

x,sec

x,csc

x及其四則運算構(gòu)成的函數(shù)的不定積分,可以通過萬能替換t

tan

x2將該函數(shù)的不定積分化為t的有理函數(shù)積分.1

tan22

tan

x

1

tan21

tan2sinx

2

cos

x

2

.

x2xx1

tan22

tan

xtan

x

2

x2x2

.x22

tan1

tan2csc

x

x2x2x2

.1

tan21

tan2sec

x

2

tan

x2Edited

by

LinGuojian12018/12/201

tan2cotx

2

.萬能公式:x2sec2

xx1

tan2dt.21

t

22

2sin

x

2cos

x2cos2x

2

tan

cosxsin

x

2

sin

cos

22

2sec2

2x22

tan

xxx2

2x21

tan22

tan

x

2

.21

tan2x1

tan2

2

2

.x2x

]22x

[1

tan2cos2

xsin

2

xx

x

xcos

xcos2

sin

2

cos2

[1

2

]

cos2222令t

tan

x

,則x

arctan

t

x

2

arctan

t.2

2故dx

2tEdited

by

LinGuojian22018/12/20,

cot

x

1

t

2

.2t

2t,

cos

x

1

t

2

,

tan

x

1

t

2

1

t

2

1

t

2且sin

x

5

cos

x例:

1

dx.1

t

2dt,且cos

x

1

t

.2211

t

2dt

dx

2t

2

t

26

4t1

dtt

221

t25

1

t

21解:令t

tan

x

,則x

2

arctan

t

dx

22

5

cos

x1故66

arctan(6

tan

x

)

C.3

232

arctan(2

tan

x

)

C

3

22

12t

CEdited

by

LinGuojian32018/12/20t

2dt

1

332

arctan2dt

13211

3

2t

22.

sin

2x(7

tan2

x)dx例:dt.11

t

2解:令t

tan

x,則x

arctan

t.故dx

2t1

t

2dx

1(7

t

2

)dt

1 1

t

2

11

t

2

2t(7

t

2

)

1

t

2dt

sin

2x(7

tan2

x)11d

(7

t

2

)14

28 7

t)dt

1

ln

t

11

tdt

1

2(

2t(7

t

2

) 14

t

7

t

2

C

1

ln

tan

x

1

ln

7

(tan

x)2

C.14

28Edited

by

LinGuojian42018/12/20

1

ln

t

1

ln

7

t

214

282

sin

x

cos

x

5例:

1

dx.1

t

2,

cos

x

1

t

2

.2t1

t

221

t

22

2解:令t

tan

x

,則x

arctan

t

x

2

arctan

t.11dt.且sin

x

2dt

1

5故dx

21

t212t

1

t

22dx

2sin

x

cos

x

5dtt

24t

1

t

5t2

5t

2t

212t

23

tan

x

1

1

arctan

2

C.5

5t

1

1

d

(t

1)

1

3

arctan 3

C

1

arctan

3t

1

C5

)2(t

1)2

(dt

Edited

by

LinGuojian52018/12/20

4t

4

6t

2dt13

dt

1t

2

t

23

3(t

1)

53

913

dt

1

2t

21222

3t

2.dx

7

tan2

x例:

7

tan2

x

(7

t

2

)(1

t

2

)dx

dtdt

161t

1

7

t

21276

7

1

arctan

t

1

arctan

t

C76

7

1

x

1

arctan

tan

x

C

.dt.Edited

by

LinGuojian62018/12/201

t

21解:令t

tan

x,則x

arctan

t.故dx

有些函數(shù)不定積分雖然存在,但不能用初等函數(shù)表示,即不可積.sin

x

dx,xcos

x

dx,x

1

ln

x

x2

e

dx,dx,x2ex

e

dx,

x

dx,下列函數(shù)不定積分存在但不能初等函數(shù)表示:結(jié)論:任何續(xù)函數(shù)都存在不定積分.Edited

by

LinGuojian72018/12/20Edited

by

LinGuojian82018/12/20ln

x例:

[ln

ln

x

1

]dx.

x

ln

ln

x

x

1

1

dx

1 dxln

x

x

ln

x

x

ln

ln

x

1

dx

1

dx

x

ln

ln

x

C.ln

x

ln

x有些函數(shù)不定積分中,某一項不能用初等函數(shù)表示,即不可積.但對另外項的積分可能產(chǎn)生與該不可積項符號相反的項,從而相互抵消.ln

x解:

[ln

ln

x

1

]dx

ln

ln

xdx

1

dxln

x

ln

x

x

ln

ln

x

xd

ln

ln

x

1

dx例:x2x

cos

x

sin

x

dx.

x2sin

x

dxx

x2x

cos

x

sin

x

dx

cos

x

dx

解:

cos

x

dx

sin

xd

1x

x1

d

sin

x

cos

x

1

cos

x

1cos

x

dx

1

sin

x

C.xEdited

by

LinGuojian92018/12/20ex

(1

sin

x)

dx.1

cos

x例:xexex1

cos

xdx

1

cos

xdx

dx

1

cos

x

1

cos

x1

cos

xdx

deex

(1

sin

x)ex

sin

xsin

x解:1

cos

xxdx

e

sin

x

C.1

cos

x1

cos

xxdx

e

sin

x

1

cos

xexex

x

cos

x

cos2

x

sin2

x(1

cos

x)2dxex

sin

x

1

cos

x

dx

1

cos

x

eex(1

cos

x)2dxex

sin

x

x

cos

x(1

cos

x)

sin

x(

sin

x)

1

cos

x

dx

1

cos

x

eex

xdx

e

sin

x

exd

(

sin

x

)1

cos

x

1

cos

xex1

cos

xEdited

by

LinGuojian102018/12/20ex1

cos

xdx

ex

sin

x1

cos

x

e

x1

cos

x(1

cos

x)2dxxex例:

x

12dx.ex

exdx

x

1

dx

(x

1)2

dx(x

1)2xex(x

11)ex解:I

(x

1)2

dx

x

1x

1ex

dx

C.

e

dx

ex

1

x

1

x

1exexEdited

by

LinGuojian11/12/20

dx

ex

1

x

1

x

1

dex

1

1 x

1

xx

e

dx

exdx

1xxex例:x

12

dx.解:

(x

1)21x

1dx

xexdxexx

1

C.

x

1

exdxexx

1xx

1

(x1)exdx

x

exx

1exx

ex

xexdxx

1x

1x

1

1

x

d

(xe

)

x

x

ex

1Edited

by

LinGuojian122018/12/20cost2

21

u

21

u2

arcsin

u

u

C故

(1

x)arcsin(1

x)dx

td

cos

t

t

cos

t

cos

tdt

t

cos

t

sin

t

C.由于cos

t

1

sin2

t

1

u

2

且t

arcsin

u例:

(1

x)

arcsin(1

x)dx

.2x

x2解:

(1

x)arcsin(1

x)dx

(1

x)arcsin(1

x)dx.2x

x2

1

(1

x)2令u

1

x,則dx

du.故

(1

x)arcsin(1

x)dx

u

arcsin

udu.1

(1

x)2

1

u

2再令u

sin

t,

t

.則

u

arcsin

udu

t

sin

t

cos

tdt

t

sin

tdt2x

x2 1

(1

x)2

arcsin(1

x)

(1

x)

C

2x

x2

arcsin(1

x)

x

C.Edited

by

LinGuojian132018/12/20xdx

.例:

arctanx

(1

x)解:

令u

x

,則x

u2

dx

2udu.故

arctan

x

dx

2u

arctan

u

du

2

arctan

u

duu(1

u2

) 1

u2

2

arctan

ud

arctan

ux

(1

x)

(arctan

u)2

C

(arctanx

)2

C.Edited

by

LinGuojian142018/12/20ln

x

1例:

(ln

x)2

dx.

ln

x

(ln

x)2dx

(ln

x)2ln

x

1

dx

1dx1解:

1

dx

xd

1

ln

x

ln

x

1

dx

x

1

1

dxln

x

ln

x

ln

x

C.Edited

by

LinGuojian152018/12/20ln

xxln

x

1例:

(ln

x)2

dx.解:

令t

ln

x

x

et

dx

et

dt.

C.ln

xln

x

et

1

C

eln

x

tx

C

1etdt

et

d

( )

dt

et

det1

1t

tt

ttetettEdited

by

LinGuojian162018/12/20etdt

t2dt

dtt

2t2dx

et

dt

ln

x

1

t

1tet

et故

(ln

x)2t1tt

tt

e

dt

et

1

e

dt11例:

1

x2

arctan

xdx.x1111

t則1

x2arctanxdx

2arctan

tdt解:令t

1

,1Edited

by

LinGuojian172018/12/202

arctan

td

arctan

t

1

(arctan

t)2

C22

x

1

(arctan

)

C.dx

(x

ln

x)21

ln

x例:dx

(x

ln

x)21

xdx

x

ln

x

(x

ln

x)21

ln

xdx

x

ln

x

1

xdx

1解:x

(x

ln

x)21

ln

xdx

2ln

x1

ln

x2x

(1

)另解:

C.x

lnx1x

lnx11xdx

x

lnxdx

xx

lnx

(x

ln

x)211x

ln

xx

ln

xdx

xdx)ln

x1

d

ln

x2xdx

(1

C.Edited

by

LinGuojian182018/12/20x

lnx11x

C

xxx1

ln

xd

(1

ln

x

)

(1

ln

x

)2

dx.x

2

ex例:

x

22(x

2)2dxx2ex[(x

2)2

4(x

2)

4]ex解:I

(x

2)2

dx

dx

4

exdx

4exx

2(x

2)2dxex

ex

4dx

4ex

4

dx

1

exx

2

x

2exx

2

ex

4x

2

1

x

2dx

4

exdex

C.Edited

by

LinGuojian192018/12/20x

21

ex

4exdx.x

2

ex例:

x

221x

2dx

x

e

d解:2

x2(x

2)x2ex

2xe

x

x2ex

2xdxx

2x

2

x

22

xd

(x2ex

)

x

e

x2ex

1

1x

22

x

xdex

x

e

xex

exdxx

22

x

x

e

xex

dxx

2x

e

x(x

2)ex

e2

xxdx

2

xx

2

x

2

xex

ex

C.Edited

by

LinGuojian202018/12/20x

22

x

x

ex3

3

x3

4例:

1

dx

.t

tdx

(

1

)dtt13

3

3x

x

4故t

21

1t

2(

)3

4(

)3

31解:令x

1

dx

dt

.

11

1

12

1

123d

(4t

1)34t3

13d

(4t

1)

34t3

1

t

dt

dtt

2t34t3

134t3

1

C.Edited

by

LinGuojian212018/12/2012

21

32x31

4

x32

(4t3

1)3

C

[

]38

dx

.1

3例:x2

x2dt

dt

dx

故

d

t

.t

21解:令x

1

dx

tdt.1(1)2

3t1

(

t12

)dt

(1)2

(1)2

3t

t13t2

13t2

1x2

x2

3tt13x3

x113133(1

)2

1

C

x

132x

3

C.2x

3

C

x2x2

323t

1

C

C

3

C.13x1

3x2dx

x2

x21

1dt

(2):當t

0時,d(3t2

1)dt

3t2

13t2

163t2

1tt13x133(13

132x

3

C.x2x2

321x)

1

C

23t

1

C

C