基本積分公式表（課堂PPT）

1、.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 .2 dxx211)4(Cx arctan dxx211)5(Cx arcsin xdxcos)6(Cx sin xdxsin)7(Cx cos xdx2sec )8(Cx tan xdx2csc )9(Cx cot.3 xdxxtansec)10(Cx sec xdxxcotcsc)11(Cx csc dxex)12(Cex dxax)13(Caax ln.4第二節(jié)第二節(jié) 換元積分法（一）換元積分法（一）.5一、第一一、第一換元積分法換元積分法問題問題 d

2、xex2? ,2數(shù)數(shù)不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？.6 dxex2 xe2)2( xd21 21 xe2)2( xd xu2 21 ue du 21ueC 21xe2C .7一般情況下：一般情況下：, )( )( uFuf有原函數(shù)有原函數(shù)設設)()( ufuF 即即 )( duuf uF)(C 可導可導若若)(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

3、 C )(uFC duuf)(:,我我們們就就得得到到下下面面的的定定理理這這樣樣.8設設)(uf具具有有原原函函數(shù)數(shù)，)(xu 可可導導，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.9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 )2c

4、os(x 21C )2cos( x21 C .10解解2 xdx2sin xxcossin2dx xsin)(sinxd2 xusin 令令 u du2 22u2C 2uC x2sinC .11解解3 xdx2sin xxcossin2dx xcos)(cosxd2 xucos 令令 u du2 22u2 C 2u C x2cos C xxsincos2dx) 1( xcos)(cosxd2 .12例例2 dxx231 x231 )23(xd xu23 令令 u1du21 |ln u21C |23|ln21x C 21 x231 )23(xd 21.13例例3 dxx)43(sec2 )43(

5、sec2 x)43( xd 43 xu令令 u2secdu31 utan31C )43tan(31 xC 31 )43(sec2 x)43( xd31.14例例4 dxxx21 21x )1 (2xd 21xu 令令 udu21 21 21 21x )1 (2xd 21 2332uC Cu 2331 232)1(31x C .15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.16例例6)0( 122 adxxa 1 2a

6、 (221ax )dx 1 21a221ax dx 1 21a2)(1ax )(axda 1 a12)(1ax )(axd axarctana1C .17例例7)0( 122 adxxa )1( 1 222axa dx 1 a1221ax dx 1 a12)(1ax )(axda 1 2)(1ax )(axd axarcsinC .18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 |lna

7、xax a21C .19例例9 dxxx)ln21( 1 xln211 )ln21 (xd |ln21|ln21x C 21 xln211 )ln21 (xd 21.20例例10 dxxex3 xe3)3(xd 32xe332 xe3)3(xd32C .21例例11 sin3xdx xxsinsin2 dx )(cosxd )cos1(2x )cos31(cos3xx C Cxx 3cos31cos.22例例12 cossin52xdxx x2sindx 22)sin1 (x )(sin xdx2sin x2sin)sinsin21(42xx )(sin xd )sinsin2(sin642x

8、xx )(sinxd 3sin3xx5sin52 7sin7x C xxcoscos4.23例例13 cos2xdx dx dx)22cos21(x dx21 dxx22cos 2x dxx 2cos21 2x )2(2cos xxd4122cos1x Cxx 42sin2類似可求類似可求 cos4xdx.24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 413

9、21 Cxxx 4sin3212sin4183.25例例15 csc xdx dx dx2cos2sin21xxxsin1 dx2cos2cos2sin212xxx.26 dx2cos2tan212xx dx2sec2tan212xx )2(xd2sec2tan12xx )2(tanxd2tan1x |2tan|lnxC .272tanx 2cos2sinxx 2cos2sinxx2sin22sin2xx xsin2sin22x xsinxcos1 xxsincos xsin1 xcot xcsc xdxcscCx |2tan|lnCxx |cotcsc|ln.28例例16 xdxsec dx

10、xcos1 dxx)2sin(1 )2()2sin(1 xdx 例例15Cxx | )2cot()2csc(|ln Cxx |tansec|ln.29例例17 xdx6sec xdxx24secsec ) (dx4secxtan 22)tan1(x )(tanxd )tantan21(42xx )(tanxd xtan x3tan32 5tan5xC .30例例18 xdxx35sectan xdxxxxtansecsectan24 ) (d22)1(sec xxsec xxx224sec)1sec2(sec )(sec xd )secsec2(sec246xxx )(sec xd x7sec

11、71 x5sec52 x3sec31C x2sec.31例例19 xdxx2cos3cos )cos5(cos21xx dx 21dxxx)cos5 (cos (21xdx5 cos xdxcos) 5121 )5(5 cosxxd xdxcos 51(21x5sin xsin) 101x5sin xsin21C C .32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.33例例2121 求求.11dxex 解解dxex 11dxeeexxx 11dxee

12、xx 11dxeedxxx 1)1(11xxededx .)1ln(Cexx .34例例2222 求求.)11(12dxexxx 解解,1112xxx dxexxx 12)11()1(1xxdexx .1Cexx .35例例2323 求求.12321dxxx 原式原式 dxxxxxxx 123212321232dxxdxx 12413241)12(1281)32(3281 xdxxdx .12121321212323Cxx .36例例2424 求求解解.cos11 dxx dxxcos11 dxxxxcos1cos1cos1 dxxx2cos1cos1 dxxx2sincos1 )(sinsin1sin122xdxdxx.sin1cotCxx dxxxx)sincossin1(22.37例例2424 求求另解另解.cos11 dxx dxxcos11 dxx2cos212 )2(2cos12xdx )2(2sec2xdxCx 2tan.38解解例例2525 設設 求求 .,cos)(sin22xxf )(xf令令xu2sin ,1cos2u