定積分的換元法和分部積分法

1、1第四節(jié)第四節(jié) 定積分的換元法定積分的換元法., 積積分分限限要要做做相相應(yīng)應(yīng)地地變變動(dòng)動(dòng)用用換換元元法法時(shí)時(shí)注注釋釋 dtttfdxxfbattbatxbaxfba)( )()( ,)( ),(,)( )2( )( ,)( )1( )(,)( ，則則有有導(dǎo)導(dǎo)數(shù)數(shù)，上上單單值值且且具具有有連連續(xù)續(xù)或或在在；滿滿足足條條件件：函函數(shù)數(shù)上上連連續(xù)續(xù)在在區(qū)區(qū)間間設(shè)設(shè)函函數(shù)數(shù)定定理理換換元元積積分分公公式式2. )( )( xFxf所所以以存存在在原原函函數(shù)數(shù)連連續(xù)續(xù)，因因?yàn)闉樽C證).( )()( )()(ttftxfdtdxdxdFtF 知知：由由復(fù)復(fù)合合函函數(shù)數(shù)的的求求導(dǎo)導(dǎo)法法則則.)()()()

2、( )( babadxxfxFtFdtttf 故故.)( )()( dtttfdxxfba即即3;1)1( 1 102 xxdxI計(jì)計(jì)算算下下列列定定積積分分例例 .01 cos11.0 )(,)2()2(241xxxxexfdxxfIx其其中中4.4cossinln214)cossinsincos1(21cossincos)1(202020sin ttdtttttdttttItx解解.212121tan212tancos11)()2(42001200121222 eetdttedttdttfItttx5 證證明明上上連連續(xù)續(xù)在在設(shè)設(shè)例例.,)( 2aaxf ).()(0).()()(2)()

3、()(00 xfxfxfxfdxxfdxxfxfdxxfaaaa.11cos2)2(;1sin)1(31122442 dxxxxxIdxexIx求求下下列列定定積積分分例例6.418sin1sin1sin) 1 (4402022 dxxdxexexIxx解解 112112211cos112)2(dxxxxdxxxI 1021022)11(4114dxxdxxx.4144102 dxx7.)()()2(;)()()1(.),()(400 TnTaaTTaadxxfndxxfdxxfdxxfTxf證證明明為為周周期期的的連連續(xù)續(xù)函函數(shù)數(shù)上上以以在在設(shè)設(shè)例例.)()()()()()()()()()(

4、)() 1(00000000 TaTaTaTTaTaaaatTxTaTdxxfdttfdxxfdxxfdxxfdxxfdxxfdxxfdttfdttTfdxxf由由證證8.)()()()0()()(0)()()( )()(000 TTaaTxTxxdttfdttfdttfFCCxFxfTxfxFdttfxF令令常常數(shù)數(shù)或或令令.)()()()()()()()()()2(0000)1(200 TTTTnTTnTTTnTnTaadxxfndxxfdxxfdxxfdxxfdxxfdxxfdxxfdxxf9.)1(tan2sin)2(;2cos1)1(520 aandxxxIdxxI求求下下列列定定積

5、積分分例例.22cos22cos2cos2cos2cos2)1(20220002ndxxndxxndxxndxxdxxInn 解解10.224cos122sin)1(tan2sin)1(tan2sin)1(tan2sin)()2(20222222022 dxxxdxdxxxdxxxIxxxf為為周周期期以以被被積積函函數(shù)數(shù)11 .)tan1ln(.)()(21)(, 0)(64000 dxxIdxxafxfdxxfaxfaxata計(jì)計(jì)算算并并由由此此證證明明上上連連續(xù)續(xù)在在設(shè)設(shè)例例.cos1sin.)(sin2)(sin.)(70200 dxxxxIdxxfdxxxfxfxt由由此此計(jì)計(jì)算算并

6、并證證明明為為連連續(xù)續(xù)函函數(shù)數(shù)設(shè)設(shè)例例12. 2ln82ln21)tan1tan11)(tan1(ln21)4tan(1ln()tan1ln(21404040 dxdxxxxdxxxI解解.4)arctan(cos2)(coscos112cos1sin2200202 xxdxdxxxI解解13第五節(jié)第五節(jié) 定積分的分部積分法定積分的分部積分法.)(,)(),( bababavdxuuvdxuvvuuvuvbaxvxu則則上上具具有有連連續(xù)續(xù)導(dǎo)導(dǎo)數(shù)數(shù)，在在區(qū)區(qū)間間設(shè)設(shè). b baabavduuvudv即即分分部部積積分分公公式式14 10;11arctan)1( 1dxxxI計(jì)計(jì)算算下下列列定定

7、積積分分例例.)(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?yàn)闉?6 11211)()()3(dxeexdxeexIxxxx.4)(2)(2)(2)(2)(2)(2110110101010 eeeeedxeeeexeexddxeexxxxxxxxxxx

8、17.)(,)(21010)2( dxxfIdtexfxtt求求設(shè)設(shè)例例).0(,5sin)()(,2)(,)(30fdxxxfxffxf求求且且有有二二階階連連續(xù)續(xù)導(dǎo)導(dǎo)數(shù)數(shù)設(shè)設(shè)例例 ).12(3254231).2(22143231cossin)1(420202knnnnnknnnnnxdxxdxIWallisnnnn公公式式證證明明題題例例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 證證21 .41)( ()()(;)( )()(.1)(,0)1(,1 ,0)()2(102102210102 dxxfdxxfxiidxxfxxfIidxxffxf證證明明求求且且上上有有連連續(xù)續(xù)導(dǎo)導(dǎo)數(shù)數(shù)在在閉閉區(qū)區(qū)間間設(shè)設(shè)函函數(shù)數(shù) . 0)()(,), 0(

10、. 0cos)(, 0)(, 0)()3(212100 ffdxxxfdxxfxf使使個(gè)個(gè)不不同同的的點(diǎn)點(diǎn)兩兩內(nèi)內(nèi)至至少少有有在在證證明明且且上上連連續(xù)續(xù)在在設(shè)設(shè)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則則若若由由且且內(nèi)可導(dǎo)內(nèi)可導(dǎo)在在上連續(xù)上連續(xù)在在則則令令證證24.sin)3(.)sin()2(.)1()1(.10021122 nxdxxIdttxdxddxxxI填填空空題題練練 習(xí)習(xí) 題題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計(jì)計(jì)算算下下列列定定積積分分的的奇奇偶偶性性如如何何？為為奇奇函函數(shù)數(shù)，問問連連續(xù)續(xù)且且上上在在的的導(dǎo)導(dǎo)函函數(shù)數(shù)設(shè)設(shè)函函數(shù)數(shù))(),()( )(. 2xfyxfxfy ).()(0)( )()(),(.)(xfxfdttfxfxfxxfyxx 由由一一定定為為偶偶函函數(shù)數(shù)解解27.2)cos2(2)sinsin(2)(sin2cos2).20()2(20202020202 ttdtttttdtdttIttx則則令令.42)(2)(22)1(1101010101111 edxexeexddxxedxexdxexIxxxxxx解解28.3,2122.)1(.51102 nInnIdxxInnnn明明證證設(shè)設(shè).)(,cos1)(.4200 dxxfIxdttxftx求求設(shè)設(shè)29. 12sinsin)(sin)()()(cos1)()()()()()(