-2hx空.-aw =片=6g 例5.设x=n+,求∫/x. 解，令1=nx,则x=d,0=+ 蝌=螂+e-e+r2 e0*ero-e0*er0*rc =x-(l+")In(l+e)+c 例6。∫m。本注此想男解法见第节到4) nf高小 +「1 24am2+∫soc25d 2tan =htm+m芝+m+c 鞋三原式j+,k-2+小∫在-∫ sin3 x -∫cse2h-cox+2 ss 11 而 ∫csc2t=-∫cscxdcotx=-cscxcotx-∫cot2 xcsexdx=-cscxcotx-∫csc.x(csc2x-l)d =-cscxcot.x-∫csc3xt+「csc.xdx=-cscxcotx-+Incscx-cot-「csc2xt .fese xd =-csexcotx+nescx-cot+c 时n0o可-分oa+兮oecm时-c+2x+C 例7.∫3n2x+2sn dx 心5 ＝ 3 2 3 2 ln ln 1 1 ln 3ln 1 3 3 2ln 6 ln ( ) x x x x x dx xd x x x x x x x − − +   = − − −   ＝ 3 2 2 ln 3ln ln 1 6( ) x x x dx x x x x − − − −  ＝ 3 2 ln 3ln 6ln 6 x x x C x x x x − − − − + 例５．设 ln(1 ) (ln ) x f x x + = ，求  f (x)dx ． 解． 令 t x = ln ，则 ln(1 ) , ( ) t t t e x e f t e + = = ln(1 ) 1 ( ) ln(1 ) ( ) ln(1 ) 1 x x x x x x x e f x dx dx e d e e e dx e e + - - = = - + = - + + 蝌 蝌 + ln(1 ) (1 ) ln(1 ) ln(1 ) 1 x x x x x x x e e e dx e e x e c e - - = - + + - = - + + - + + ò + (1 )ln(1 ) x x x e e c - = - + + + 例６．  + + dx x x x sin (1 cos ) 1 sin (注：此题另一解法见第四节例４) 解法一 原式=     = + + + + dx x dx x x x dx x x dx 2 2 cos 1 2 cos 2 4sin 1 sin (1 cos ) 1 cos 3 2 =     + + + = ) 2 ( 2 ) sec 2 (tan 2 2 tan 2 1 tan ) 2 ( 2 cos 1 ) 2 ( 2 cos 2 2sin 1 2 2 3 2 x d x x d x x x d x x d x x = C x x x + + + 2 tan 2 tan 4 1 2 ln tan 2 1 2 解法二 原式=  = + − dx x x x 3 sin (1 sin )(1 cos )     + − − dx x x dx x x xdx xdx 3 2 3 2 sin cos sin cos csc csc = x x xdx x sin 1 2sin 1 csc cot 2 3 − + +  而     csc xdx = − cscxd cot x = −cscx cot x − cot x cscxdx = −cscx cot x − cscx(csc x −1)dx 3 2 2 =    − x x − xdx + xdx = − x x + x − x − xdx 3 3 csc cot csc csc csc cot ln csc cot csc   xdx = − x x + ln cscx − cot x + C 2 1 csc cot 2 1 csc3 故 C x x dx x x x x x x x x = − + − − + + + + +  sin 1 2sin 1 ln csc cot cot 2 1 csc cot 2 1 sin (1 cos ) 1 sin 2 例７． 1 sin 2 2sin dx x x +  解法一 原式= 2 2 2 3 sin cos 2 2 2sin (cos 1) 4sin cos 8sin cos 2 2 2 x x dx dx dx x x x x x x + = = +   