微积分II课程63习题解答

1、练习6.31. 用第一换元法求下列不定积分(1)8(5x - 3) dx;解:(2)(5x _3)8dx = 1 (5x _3)8d(5x _3)5 L=丄(5x - 3)9 C。45x 2x2 -1dx ；解:.2x2 -1dx = -. 2x2 -1d(2x2 -1)4 1 225(2x -1)2 - C4 3= (2x2 -1)3 C。6(3)解:ln-x dx =、1 In xd(1 ln x)x(4)=j J(1_ln x)3 C。edx;解:edx = _ d(e)-eC。(5) cscxdx ；解:1cscxdx=dx sin xsin x ,2 dxsin x_12 d cos

2、xsin x1 .2dcosxcos x T11/1 1()d cosx2 cosx -1 cosx 12(l ncosx1 In cosx+1) +Clln2COSX -1COSX 1In cotx cscx +C .(6)sin2 xcos4 1 xdx;解:；24sin x cosxdx= sin2 x cos x cos2 xdx2sin 2x cos2x 1 , dx21 2sin 2x(cos2x 1)dx122(sin 2x cos2xdx 亠 isin 2xdx) 8dx1. 21 1-cos4xsin 2xd sin2x16 8 2丄 x 丄 sin 4x 丄 sinx 2x

3、 C .166448(7)解:-1 1 dx x 1(8)解:(9)1 2 . ,= 一x x + 1 n x +2dx;x2 1x只1 二 dx=2x212 x211 2=-ln(x21) C.二 1 dx;x2 2x 11 C.1 2d(x2 1)12x 2x1解:r 1dx =2dx(x+1)1(10)dx;x2 +2x31 1111解：dx =dxdxx2+2x34x+34x1(11)x22x 5dx;2、“ 厂2 -21 n(1. x 2) C .解:1(x 1)2 - 22dx1 丄X1arctanC .2 2(12)1sin2x 3cos2xdx ;解:2dx =sin x 3c

4、os xtan2 x 3d tanxarcta n3tan x(13)已知 F (u)二 f (u)，求 f(ax b)dx (a = 0).1解：f (ax b)dx= f (ax b) d (ax b)a L1=F (ax b) C .a2. 用第二换元法求下列不定积分.(1)1 dx;1. x 2解：令.x 2 =t ,12tdxdt1.x 21 t=2dt2dt十t=2t -21 n(1 t) Cx3(2)dx ；j+X4x3解：r1 3CCT 1 111_ d(1 x4)，令 31一x4 二 t ,dx= 11 3 1 x444(t2-t In |1 t|) C4 2一331 X4-

5、In(31 X41) C .44(3) i .= dx ；.(1-x2)3解：令 x = si ntdx-(1-x2)3-Sdtcos t=seEtdttant C二2 C.1(4)dx ；X2 .3 X2,亦 seStdx=dt3tan2t 3 3tan2t仝琴dt3ta n2t卑dt3si n2tC3sin t3x(5).dx ;1 9x2解:.1 9x2dx=1dx3【9 x2= 1ln3=1|n(3x . 1 9x sin4t) C。(6)2l r22a -xdx (a 0);解：令 x = asint-22 dx =.a -xa2 sin2t a costdtacosta22/ ta(2a2 sin 2tdt1 - cos2t 亠2