定积分的换元法和分部积分法

1、1第四节第四节 定积分的换元法定积分的换元法., 积积分分限限要要做做相相应应地地变变动动用用换换元元法法时时注注释释 dtttfdxxfbattbatxbaxfba)( )()( ,)( ),(,)( )2( )( ,)( )1( )(,)( ，则则有有导导数数，上上单单值值且且具具有有连连续续或或在在；满满足足条条件件：函函数数上上连连续续在在区区间间设设函函数数定定理理换换元元积积分分公公式式2. )( )( xFxf所所以以存存在在原原函函数数连连续续，因因为为证证).( )()( )()(ttftxfdtdxdxdFtF 知知：由由复复合合函函数数的的求求导导法法则则.)()()()

2、( )( babadxxfxFtFdtttf 故故.)( )()( dtttfdxxfba即即3;1)1( 1 102 xxdxI计计算算下下列列定定积积分分例例 .01 cos11.0 )(,)2()2(241xxxxexfdxxfIx其其中中4.4cossinln214)cossinsincos1(21cossincos)1(202020sin ttdtttttdttttItx解解.212121tan212tancos11)()2(42001200121222 eetdttedttdttfItttx5 证证明明上上连连续续在在设设例例.,)( 2aaxf ).()(0).()()(2)()

3、()(00 xfxfxfxfdxxfdxxfxfdxxfaaaa.11cos2)2(;1sin)1(31122442 dxxxxxIdxexIx求求下下列列定定积积分分例例6.418sin1sin1sin) 1 (4402022 dxxdxexexIxx解解 112112211cos112)2(dxxxxdxxxI 1021022)11(4114dxxdxxx.4144102 dxx7.)()()2(;)()()1(.),()(400 TnTaaTTaadxxfndxxfdxxfdxxfTxf证证明明为为周周期期的的连连续续函函数数上上以以在在设设例例.)()()()()()()()()()(

4、)() 1(00000000 TaTaTaTTaTaaaatTxTaTdxxfdttfdxxfdxxfdxxfdxxfdxxfdxxfdttfdttTfdxxf由由证证8.)()()()0()()(0)()()( )()(000 TTaaTxTxxdttfdttfdttfFCCxFxfTxfxFdttfxF令令常常数数或或令令.)()()()()()()()()()2(0000)1(200 TTTTnTTnTTTnTnTaadxxfndxxfdxxfdxxfdxxfdxxfdxxfdxxfdxxf9.)1(tan2sin)2(;2cos1)1(520 aandxxxIdxxI求求下下列列定定积

5、积分分例例.22cos22cos2cos2cos2cos2)1(20220002ndxxndxxndxxndxxdxxInn 解解10.224cos122sin)1(tan2sin)1(tan2sin)1(tan2sin)()2(20222222022 dxxxdxdxxxdxxxIxxxf为为周周期期以以被被积积函函数数11 .)tan1ln(.)()(21)(, 0)(64000 dxxIdxxafxfdxxfaxfaxata计计算算并并由由此此证证明明上上连连续续在在设设例例.cos1sin.)(sin2)(sin.)(70200 dxxxxIdxxfdxxxfxfxt由由此此计计算算并

6、并证证明明为为连连续续函函数数设设例例12. 2ln82ln21)tan1tan11)(tan1(ln21)4tan(1ln()tan1ln(21404040 dxdxxxxdxxxI解解.4)arctan(cos2)(coscos112cos1sin2200202 xxdxdxxxI解解13第五节第五节 定积分的分部积分法定积分的分部积分法.)(,)(),( bababavdxuuvdxuvvuuvuvbaxvxu则则上上具具有有连连续续导导数数，在在区区间间设设. b baabavduuvudv即即分分部部积积分分公公式式14 10;11arctan)1( 1dxxxI计计算算下下列列定定

7、积积分分例例.)(1()3(;)2cos1()2(110 dxeexxIdxxeIxxx.211211)1(4112111arctan)1(102102210210 xxxddxxxxxxI解解15).1(54)2cos1, 1),1(512cos2cos41)(2sin21)(2cos2cos)2(0000000 edxxeIedxeexdxexdxeeexdeexdxdxexxxxxxx（所所以以因因为为16 11211)()()3(dxeexdxeexIxxxx.4)(2)(2)(2)(2)(2)(2110110101010 eeeeedxeeeexeexddxeexxxxxxxxxxx

8、17.)(,)(21010)2( dxxfIdtexfxtt求求设设例例).0(,5sin)()(,2)(,)(30fdxxxfxffxf求求且且有有二二阶阶连连续续导导数数设设例例 ).12(3254231).2(22143231cossin)1(420202knnnnnknnnnnxdxxdxIWallisnnnn公公式式证证明明题题例例18).1(2121)1(21)( )()( , 0)1(1011021101101012222 eexdedxxedxxxfxxfIexffxxxx由由解解19.3)0()0(2)0()(cos)( )sin)( (cos)( )cos)()( sinc

9、os)(sin)()(50000000 ffffdxxxfxxfdxxxfxxfxdfxxdxfdxxxfxf由由解解20. 1,2. )2(1)1()1()sin1(sin)1(cossin)1(sincos)cos(sin102220222022201201 IInInnIInIndxxxnxdxxnxxxxdInnnnnnnnn 证证21 .41)( ()()(;)( )()(.1)(,0)1(,1 ,0)()2(102102210102 dxxfdxxfxiidxxfxxfIidxxffxf证证明明求求且且上上有有连连续续导导数数在在闭闭区区间间设设函函数数 . 0)()(,), 0(

10、. 0cos)(, 0)(, 0)()3(212100 ffdxxxfdxxfxf使使个个不不同同的的点点两两内内至至少少有有在在证证明明且且上上连连续续在在设设22.41)( )()( ()()(.21)(21)(21)(21)()()(210102102210210210210 dxxfxxfdxxfdxxfxCauchyiidxxfxxfxfxdxfdxxfIi不不等等式式由由解解23 . 0)()(0)( )( ),(), 0(:. 0)(), 0()(0sin)(0sin),0(0)(, 0)(), 0(. 0sin)(sin)(cos)()(coscos)(. 0)(, 0)0(,

11、), 0(, 0)(,)()(2121210000000 ffFFccRollecFcdxxxFxorxFxFxxdxxFxdxxFxxFxxdFxdxxfFFxFdttfxFx使得使得定理定理由由故故使得使得矛盾矛盾而而则则若若由由且且内可导内可导在在上连续上连续在在则则令令证证24.sin)3(.)sin()2(.)1()1(.10021122 nxdxxIdttxdxddxxxI填填空空题题练练 习习 题题2n22sin x25.sinsin)(sin)sin()2(2020202xduudxdduudxddttxdxdxxtxux . 2)121()1(11222 dxxxxxI解解.

12、2sin)3(0nxdxnI 26;cos)2(;)()1(.32)2(011 dxxIdxexxIx计计算算下下列列定定积积分分的的奇奇偶偶性性如如何何？为为奇奇函函数数，问问连连续续且且上上在在的的导导函函数数设设函函数数)(),()( )(. 2xfyxfxfy ).()(0)( )()(),(.)(xfxfdttfxfxfxxfyxx 由由一一定定为为偶偶函函数数解解27.2)cos2(2)sinsin(2)(sin2cos2).20()2(20202020202 ttdtttttdtdttIttx则则令令.42)(2)(22)1(1101010101111 edxexeexddxxedxexdxexIxxxxxx解解28.3,2122.)1(.51102 nInnIdxxInnnn明明证证设设.)(,cos1)(.4200 dxxfIxdttxftx求求设设29. 12sinsin)(sin)()()(cos1)()()()()()(