1、基本積分公式表基本積分公式表 kdx)1(kxc )( 是常數(shù)是常數(shù)k dxx )2(11 x,c )1( xdx)3(|ln xc dxx211)4(cx arctan dxx211)5(cx arcsin xdxcos)6(cx sin xdxsin)7(cx cos xdx2sec )8(cx tan xdx2csc )9(cx cot xdxxtansec)10(cx sec xdxxcotcsc)11(cx csc dxex)12(cex dxax)13(caax ln第二節(jié)第二節(jié) 換元積分法（一）換元積分法（一）一、一、第一第一換元積分法換元積分法問題問題 dxex2? ,2數(shù)數(shù)不

2、不是是積積分分公公式式表表上上的的函函被被積積函函數(shù)數(shù)xe用直接積分法，求不出它的積分。用直接積分法，求不出它的積分。怎么辦？怎么辦？ dxex2 xe2)2( xd21 21 xe2)2( xd xu2 21 ue du 21uec 21xe2c 一般情況下：一般情況下：, )( )( ufuf有原函數(shù)有原函數(shù)設(shè)設(shè))()( ufuf 即即 )( duuf uf)(c 可導(dǎo)可導(dǎo)若若)(x u )(uf )(xf )(xfdxd )( )( xuf )(uf)( x )(xf )( x 的的原原函函數(shù)數(shù)是是 )( )( )(xxfxf dxxxf)( )( )(xf c )(ufc duuf)(

3、:,我我們們就就得得到到下下面面的的定定理理這這樣樣設(shè)設(shè))(uf具具有有原原函函數(shù)數(shù)，)(xu 可可導(dǎo)導(dǎo)，則有換元公式則有換元公式定理定理1 1)(xu dxxxf)( )( duuf)(:用法用法 dxxg)( )(xf )( x dx )(xf )(xd )(ufdu)(xu )(ufc )(xf c )(xu 是是積積分分公公式式表表上上的的函函數(shù)數(shù))(uf“湊湊” 微分法微分法例例1 1 求求.2sin xdx解解1 xdx2sin x2sin)2( xd21 x2sin)2( xd21 xu2 令令 usindu21 )cos(u 21c )2cos(x 21c )2cos( x21

4、 c 解解2 xdx2sin xxcossin2dx xsin)(sinxd2 xusin 令令 u du2 22u2c 2uc x2sinc 解解3 xdx2sin xxcossin2dx xcos)(cosxd2 xucos 令令 u du2 22u2 c 2u c x2cos c xxsincos2dx) 1( xcos)(cosxd2 例例2 dxx231 x231 )23(xd xu23 令令 u1du21 |ln u21c |23|ln21x c 21 x231 )23(xd 21例例3 dxx)43(sec2 )43(sec2 x)43( xd 43 xu令令 u2secdu31

5、 utan31c )43tan(31 xc 31 )43(sec2 x)43( xd31例例4 dxxx21 21x )1 (2xd 21xu 令令 udu21 21 21 21x )1 (2xd 21 2332uc cu 2331 232)1(31x c 例例5 xdxtan xxcossindx xcos1)(cosxd xucos 令令 u1du ) 1( xcos1)(cosxd |ln uc cx |cos|ln類似可得類似可得 xdxcot cx |sin|ln例例6)0( 122 adxxa 1 2a (221ax )dx 1 21a221ax dx 1 21a2)(1ax )(

6、axda 1 a12)(1ax )(axd axarctana1c 例例7)0( 122 adxxa )1( 1 222axa dx 1 a1221ax dx 1 a12)(1ax )(axda 1 2)(1ax )(axd axarcsinc 例例8)0( 122 adxax )( 1 axax dx a21)11(axax dx a21)11(axax dx a21ax 1dx a21ax 1dx a21ax 1)(axd a21ax 1)(axd |lnax a21 a21|lnax c |lnaxax a21c 例例9 dxxx)ln21( 1 xln211 )ln21 (xd |ln

7、21|ln21x c 21 xln211 )ln21 (xd 21例例10 dxxex3 xe3)3(xd 32xe332 xe3)3(xd32c 例例11 sin3xdx xxsinsin2 dx )(cosxd )cos1(2x )cos31(cos3xx c cxx 3cos31cos例例12 cossin52xdxx x2sindx 22)sin1 (x )(sin xdx2sin x2sin)sinsin21(42xx )(sin xd )sinsin2(sin642xxx )(sinxd 3sin3xx5sin52 7sin7x c xxcoscos4例例13 cos2xdx dx

8、 dx)22cos21(x dx21 dxx22cos 2x dxx 2cos21 2x )2(2cos xxd4122cos1x cxx 42sin2類似可求類似可求 cos4xdx例例14 cos4xdx dx dx)42cos22cos41(2xx 2)22cos1(x dx)24cos14122cos41(xx dx)84cos22cos83(xx dx 83dxx 4cos81dxx 2cos21 x83)4(4cos xdx )2(2cos xdx 41321 cxxx 4sin3212sin4183例例15 csc xdx dx dx2cos2sin21xxxsin1 dx2co

9、s2cos2sin212xxx dx2cos2tan212xx dx2sec2tan212xx )2(xd2sec2tan12xx )2(tanxd2tan1x |2tan|lnxc 2tanx 2cos2sinxx 2cos2sinxx2sin22sin2xx xsin2sin22x xsinxcos1 xxsincos xsin1 xcot xcsc xdxcsccx |2tan|lncxx |cotcsc|ln例例16 xdxsec dxxcos1 dxx)2sin(1 )2()2sin(1 xdx 例例15cxx | )2cot()2csc(|ln cxx |tansec|ln例例17

10、 xdx6sec xdxx24secsec ) (dx4secxtan 22)tan1(x )(tanxd )tantan21(42xx )(tanxd xtan x3tan32 5tan5xc 例例18 xdxx35sectan xdxxxxtansecsectan24 ) (d22)1(sec xxsec xxx224sec)1sec2(sec )(sec xd )secsec2(sec246xxx )(sec xd x7sec71 x5sec52 x3sec31c x2sec例例19 xdxx2cos3cos )cos5(cos21xx dx 21dxxx)cos5 (cos (21xd

11、x5 cos xdxcos) 5121 )5(5 cosxxd xdxcos 51(21x5sin xsin) 101x5sin xsin21c c 例例2020.)1(3dxxx )1(1)1(132xx c )1(xd 1)1( x 2)1(21 xc x 11 2)1(21x dxxx 3)1(11)1(1)1(132xx dx例例2121 求求.11dxex 解解dxex 11dxeeexxx 11dxeexx 11dxeedxxx 1)1(11xxededx .)1ln(cexx 例例2222 求求.)11(12dxexxx 解解,1112xxx dxexxx 12)11()1(1x

12、xdexx .1cexx 例例2323 求求.12321dxxx 原式原式 dxxxxxxx 123212321232dxxdxx 12413241)12(1281)32(3281 xdxxdx .12121321212323cxx 例例2424 求求解解.cos11 dxx dxxcos11 dxxxxcos1cos1cos1 dxxx2cos1cos1 dxxx2sincos1 )(sinsin1sin122xdxdxx.sin1cotcxx dxxxx)sincossin1(22例例2424 求求另解另解.cos11 dxx dxxcos11 dxx2cos212 )2(2cos12xdx )2(2sec2xdxcx 2tan解解例例2525 設(shè)設(shè) 求求 .,cos)(sin22xxf )(xf令令xu2sin ,1cos2