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1、【第一换元法例题】1、 (5x+7)9dx=(5x + 7)9dx=(5x+7)9d(5x + 7) = l(5x + 7)9 d(5x + 7)551(5x 7) d (5x 7)(5x 7) C (5x 7) C55 10501【汪】(5x+7)，=5,n d(5x+7)=5dx, n dx = d(5x + 7) 5991 n11 1ioIQXJ j|n x1dx=fln x d In x2、xx12-12-=In x d In x = 2 (ln x) C =*lnx) C111一d【注】【注】x(lnx)，=，= d(lnx) =-dx,=xxdtanxdx jinxdxsinxdx
2、dcosx3 (1)cosxcosx cosx cosxd cosx=-In |cosx | Ccosx=-In |cosx | Cdcosx【注】【注】(cosx) = -sin x,n d (cos x) =-sin xdx, n sin xdx =-d(cos x)3 (2)cosx .cot xdx = -dxcosxdx d sin xT- = sin x sin x二sin xd sin x=ln |sinx | C = In |sinx| Csin x【注】【注】(sin x)，= cosx, n d (sin x) =cosxdx, n cosxdx = d (sin x)1d
3、x =d(a x)4 (1) - dx =axax ax1-= -d(a x) = In |a x| C = In |a x| Ca x/、-【注】【注】(a +x) =1,n d(a +x) =dx, n dx =d(a +x)4 (2)dx =/、1】dx =x -a x -a1-=d(x - a) =ln|x-a| C =ln|x-a| Cx -ad(x-a) x a-【注】【注】(xa) =1,n d(xa)=dx, = dx=d(xa)1dx =x -a.4 (3)11222dx =2x -a 2a1 11x -aln 2ax a x ax=厂2a . x - a2,5()s d严c
4、x(secx+tan x)dsecx tanx.sec x+secxtanxsecx tanxd(tanx secx)d(tanx secx)secx tanx,1 , cosx cosx dx dsinx5(2) secxdx = dx = dx =2=cosx cos x cos x 1一sin xd sin x、二一二5 In22 sinx1 sin x 11 -sin x6(1)In | secx tanx| C secx tanxsin x 122sin x 1.-1C = ln21 -sin x1 +sin xcscx(cscx cot x)dx =cscx cot xcsc xd
5、x =d (- cot x - cscx)=- -d (cscx cot x) =-ln |cscx cotx| C cscx cotxcscx cotx2,csc x csc x cot x-dxcscx cotxcscx(cscx - cot x) csc x - cscx cot x6(12 cscxdx =cscx-cotx=cscx-cotx:=ln |cscx cotx| Cdxdxd(一cotx cscx) d(cscx- cotx)-;=j7cscx - cot x cscx - cot x1. dx.(1)dx = arcsin, 1 -x2.1-x2d x d7(2)1 -
6、dx22,a -xdxdxaadixaa,a2 -x2x=arcsin CUa8(1)dxx2= arctanx C 1 x2d.a1la j8(2)dx-2dx2xdda x1+axxa 1+ C , ( a a 0) a=arctan-a91= =厂厂 In | x -a | Tn | x a | C =43,2、2357、Sin1 sin3 *xcosxdx= sin2xcos5 x sinxdx = - sin2 xcos5 x d cosx98、5.75. cos x(1 cos x) cos x d cosx = (cos x -cos x) d cosx = -82一一 一一6x sin x.- 8sin x -C2 sin3 xcod xdx = sin3 xcod x cosxdx = sin3 xcos x dsin xsin x(1-sin x) dsinx= (sin x-2sin x sin x) dsinx =
