第04章不定积分习题详解

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习题4- x 5xx)dx(x25x2)dx2x2x2x 2

3x)2dx 2ln ln 2ln解： cot2x)dx dx(csc2x1)dx2=arcsinxcotxx

x2 10x8xdx

d lnsin2xdx1(1cosxdx1x1sinx2cos

cos2xsin2(7)cos d dx(cosxsinx)dx(7)cos cosxsincos

d

cos2xsin2

dx

)d

cos2xsin2cotxtanx

cos2xsin2

(sin2

cos2解：secx(secxtanxdxsec2xdxsecxtanxdxtanxsecx x:f(x)在(,)上连续

1x2C

xx 2

1x131x2C3

xlim(xC) lim(1x2C)，即1 1C 1

1 x1

2C31C2联立并令CC,可得C1 C1 1x2 x故max{1x}dx

2x2

1x1x21

xdy

x3dx1x44y(0)0，得C0y1x44 2 2 sinxsinxcosx， cosxcosxsin cos

2

1cos2x2

1cos2x都是sinxcosx的原函数4习题4-1(1)x2dx

d(x

+ (2)1dxx

d(lnx+ (4)sec2xdx=d(tanx+(3)exdx11(5)sinxdx=d cosx+ (6)cosxdx=1111

dx=d(arcsinx+

dx=d +(9)tanxsecxdx1

d(secx+C) (10)1 x21

=d(arctanx+(x1) dx=d(2 x+C)(12)xdx=(x1)2

+ x2 x2 dx2x2x2(x24)2C

) (x2x2x2

2d(x2ln4 解： xdx

ln55 解：2dxex )ex (e2x2e3x2)exdx(e2x2e3x2d(ex)1e3x1e4x2ex d d 1

d(3x 1arcsin3x41ln

21(3x21

1(3x 21(xlnx)2dx(xlnx)2d(xlnxxlnx

dx

lnln

ex

dx

e2x

1

dx22

1

44

1

2

12x2 1x23 （10）解：3x2dx3x2xdx2 3 211dx23 2

2329292 9

dx

dx d9 d9113 9

d(94x22xarcsin

94x2

dx

1 1dx2x

(x2)(xxx1xx3

d3

x x1 1t1sincos2(t)

11(arccosx)311(arccosx)22

d

dx 111111 dx1lncotxcsc2xdx1lncotxdcot 2sinxcos 1lncotxdlncotx1(lncotx)22 x解：arctanxdx2 xdx2arctanx x xdx (1(x)2(arctanx)2 1cos2x 12cos2xcos2解：cosxdx )dx d(1cos2x

cos2)d 4

xsin 1cosd 23xsin2xsin4x2

3sinxcos3sinxcos

3sinxcos3sinxcos

：

xdx

sin xd(sin sin n3x 311

10arccos

dx10arccosxd(arccosx)

10arccosx

11

dxarcsinxd(arcsinx)

arcsin22sincosxdxsin

d(sinx)sin

sinxsin3xcos5xdxsin2xcos5xdcosx(1cos2xcos5xdcos1cos8x1cos6x tan3xsec5xdxtan2xsec4xdsecx(sec2x1)sec4xdsec1secx7x1sec5x

dx cos

cosx21tanx6x1tan5x 解：令6xtxt6dx6t5dt t21x(13x)dxt3(1t2) dt t2 dt6t6arctant=66x6arctan6x (x2

(tan2t

sec2tdt 1dtsintC 11 d(1)

1x2dx

x(1)2

(1

d(x干2

1(1(1)2x

1x

x(1)2(1)2x

x 1 1 Cx解:x3sect,dx3secttantdtx2x29sec2t

dx t2

(sect3(tant1)C3 x29arccos3) 2 ，则

dx

4sin2

2costdt=

44x12(1cos2t)dt=2t2sintcostC2arcsinx1

4x2 解 dx 3x22x 3

2x31 d(x1)

2(3x1)2))3(x12))

3(

)2

1 解:

2

x21x

2

21 2(x)

d(x14 2ln(x1x21x3) )解: dx dx d(x)2 1x 2 arcsin5(2x1)5

习题4- xsin2xdx2xd(cos2x)2cos2x2cos2xd cos xexdxxdexxexexdxxexexx2lnxdxln

x)

xln

x

xln

xd

ln x2sinx2xdcosxx2sinx2xcosx2cosxd解：因为exsin2xdxsin2xdexexsin2xexd(sinexsin2x2excos2x4exsin2xdexsin2x2excos于是exsin2xdx

5arctanxdx3

xarctan3

x3darctan3 x3x3arctanx31x2dx 3arctanx3 1 d3

3

ln(1x2) 1cos d 2(xxcos2x)d 42xcos2xdx21xdsin2xx21xsin2x1sin2xd x21xsin2x1cos2xx x 1x

xdx2 x x 12 x dx x21x12 dx3xd 3 dx 9xdx2e3x 2e3x 解：因为coslnxdxxcoslnxxdcoslnxxcoslnxsinlnxdxcoslnxxsinlnxxdsinlnxcoslnxxsinlnxcoslnxd于是coslnxdx

习题4-

x

dx

x31

dx(x2x1)d

d xlnx13

x5x43

dx(x2x1)d

x2xx3xdx(x2x1)dx(8 3)d x3x2x8lnx4lnx

3lnx12x22x x 3x（3）(x2)(x21)2dxx2dxx21dx(x21)2d d(x2 d(x2 lnx22 x2 2x21dx2(x21)2(x21)2dlnx21ln(x21)2arctan2

2x2arctanxC2(x2 x26x211x4dx[4

]d

(x1) 4lnx2lnx1

C2lnx2(x1) d （5）x3x2x1dx(x1)(x21dx2x12x21d1lnx11ln(x21)1arctanx d 2d 3（6）解：3sin2x7cos2xutanx3 33

1

312 arctan2tanx31231311

t31

3t2dt3(t1 1)dt3t2tlnt11 1

1

1 x1 t dxt 1x(t21)(t21)dt(tx1 t lnt12arctantCt14-d（1）54xd d(x解：因为

54x

1(x dxx2a2

现在a1xx2dd54x

4x 4xln3xd现在n3，重复利用此三次，ln3xdxxln3x3xln2x6xlnx6x1(1x2)2d1 dx d(bax2于是现在a1b1，于是

dx dx arctanx(1x2 2(x2 2x2 2(x2

d2xx dx1arccosa于是现在 1，于

xx2 d

arccos1 x22xd

xx2 解：令tx1x2x22xdx (x1)21dx(t22t t21由积分表 （56）、（55）、（ x dx

8

a x

ln8

xaxx2a2dx (x2a2)33x2x2x2x2 dx

ln x2x2

2xdx

[2(x8

a2(x(x1)2(x1)2d

lnx1

C解：在积分表中查得（16）、（

axbddddd于是现在a2，b1 cos6xd

d

2 现在n6，重复利用 cos6xdx1cos5xsinx cos3xsinx15(1sin2xx) 24 e2xsin3xd eaxsinbxdx现在a2，b3

1a2

eax(asinbxbcosbx)1e2xsin3xd

1eax(2sin3x3cos3x)本章复

sin sin

的一个原函数，则d(F(x))=x

dxyf(xy2xx1y2yx21xlnxf(x=

已知f(xdxsinxxC，则exf(ex1dxsin(ex1ex11cos2 1cos2 11cos2 1 d1 cos 12cosx xtanx

cos2 dx(1 解：dx1

exdx1e

d(ex1

ln(ex1)解：2

5

dx2

dx5(1

3xdx 2(4

52x ln3ln4 解：(arcsinx)2dxxarcsin2x2arcsin d 1

lnxarcsin2x2arcsin 1xarcsin2x 1x2arcsinx 1x2darcsinxarcsin2x21x2arcsinx2x解：令t x1，则xt21，于d xx

2tdt(t2

2dtt2

(1t

t

)dtlnt1t

(1

)2dx

1

(1

)2]dx

1x2dx

(1

)2d2

2(1x21解： d (arcsinx)2d(arcsinx) 1(arcsin9999

1

dx dx d arcsin31 d(2x)1

19d(919 1(2 31arcsin2x 94x2 tan5xsec4xdxtan4xsec3xdsecx(sec2x1)2sec3xdsec(sec7x2sec5xsec3x)dsecxsec8xsec6xsec4x π,π)，于2t2 dx costdt1cost1dtt td()11 1 1 1 cos222 2sin2sin 11t

Carcsin2

2

t

tCarcsin x3ex2dx1x2dex21x2ex21ex2dx21x2ex21ex2lnln

x三、设f(xx1,0x1，求f(xd x解：f(x)在(,)上连续F(xx x

21xxC2 0x11 x2C

xlim(xC 12xC ，即 Clim(

x