1、第四节第四节 分部积分法分部积分法分部积分法积分公式分部积分法积分公式:vduuvudv事实上事实上,uvvuuv )(由由可得可得uvuvvu )(dxuvdxuvdxvu)(两边积分两边积分 vduuvudv从而从而(其中其中)(),(xvvxuu有连续的导数有连续的导数)dxvudxuvuv分部积分法求积分的思路分部积分法求积分的思路:udv dxxf)(分部积分公式就是将一个分部积分公式就是将一个不定积分的计算转化为另不定积分的计算转化为另一个不定积分的计算一个不定积分的计算.dxxex dxxx sindxxx cosdxxx lndxxx arctandxxx arcsin例例1
2、求求.dxxex 解解原式原式 xxdexxe.cexexx dxxex.)1(cexx dxexx 2322221dxexx .)1(2122cexx dxex dxxx cos例例3求求 )cos(xxd解解 原式原式解解原式原式xxcos xdxcosxxcos cx sindxxx sin例例2求求)(sin xxdxxsin dxx sincx cosxxsin dxxx ln例例4求求解解 原式原式)2(ln2xdx xxln22 dxxx122 xxln22 dxx 21cx 42xxln22 cxx)1ln2(42例例5求求.ln dxx 解解 原式原式xxln dxxx 1C
3、xxx lndxx lnCxx )1(ln练习练习 求求.ln3dxxx 解解 原式原式 )41(ln4xxdxx ln414 dxx 341cx 4161dxxx 1414xx ln414 xx ln414 cxx)1ln4(1614例例6求求.arcsin dxxx 解解 原式原式 )21(arcsin2xxdxx arcsin212 dxxxxx 222121arcsin21dxxx22112dxxxxx 22211121arcsin21dxxxxx )111(21arcsin21222xx arcsin212 dxx 2121dxx 21121xx arcsin212 cxaxaxa2
4、222arcsin2dxxa22)12arcsin21(212xxxcx arcsin21xx arcsin212.14arcsin412cxxx 例例6求求.arcsin dxxx 解解 原式原式 )21(arcsin2xxddxxxxx222121arcsin21令令txsin 则则tdtdxcos dxxx221 tdtttcoscossin2dtt 2sindtt 22cos1ctt 2sin4121cttt cossin2121cxxx 2121arcsin21原式原式dxxxxx 222121arcsin21xx arcsin212.141arcsin412cxxx 例例7求求.a
5、rcsin dxx 解解 原式原式xxarcsin dxxx 21xxarcsin xxarcsin 221)1(21xxdcx 21设设xxxfsin)(sin2 dxxfxx)(1 求求解解令令tx 2sin则则tttfarcsin)(dxxfxx)(1 dxxxxxarcsin1 (2002考研试题考研试题)dxxxxxarcsin1 dxxx 1arcsinxdx 1arcsin2xx arcsin12 dxxxx211112 xx arcsin12 cx 2xx arcsin12 dxx1 xx arcsin12 dxxxx211112 例例8求求解解 原式原式dxxx arctan
6、 )2(arctan2xxddxxx 22121xx arctan212 dxxx 2211121xx arctan212 dxx )111(212xx arctan212 dxx )111(212xx arctan212 cxx )arctan(21xx arctan212 cx 21xxarctan)1(212 例例9求求.arctan dxx 解解 原式原式xxarctan dxxx 21xxarctan Cx)1ln(212xxarctan)1(112122xdx 分部积分法通常解决的积分类型分部积分法通常解决的积分类型:dxexPkx)(dxkxxPsin)(dxxxP)(ln)(d
7、xxxParcsin)(uxP)(对数函数对数函数=u反三角函数反三角函数=udxkxxPcos)(dxxxParctan)(大大多为幂多为幂函数或函数或多项式多项式.)(xP例例10 求求解解 原式原式 xdxex4313xex32)12(31 xex32)12(31 xex32)12(31 )31(343xexdxxe394 dxex394xxe394 xe3274 Cxexx32)131218(271 C.)12(32dxexx )31()12(32xedxxex32)12(31 例例11 求求.3cos)12(dxxx 解解 原式原式 )3sin31()12(xdxdxx 23sin3
8、1.3cos92cx xx3sin)12(31 xx3sin)12(31 例例12 求求.arctan2dxxx 解解 原式原式 )31(arctan3xxddxxx 23131xx arctan313 xx arctan313 dxxxx )1(312dxxxxx 23131xx arctan313 dxxxxxx )1(31arctan3123cxxxx )1ln(6161arctan31223xx arctan313 dxxx 2131dxx 31261x xx arctan313 221)1(2131xxd dxxxex 21 xdxex11 xxex1)(11xxedx xxex1
9、dxex xxex1Cex )(Cxex 1例例13求求解解 原式原式dxx2)1(1xd11)(xxexex)1(xxexex11xex)1(xe例例14 求求.tanlnsindxxx 解解 原式原式)cos(tanlnxxd dxxxxxx 2seccotcostanlncosdxxxx csctanlncos.cotcsclntanlncoscxxxx 例例15 求求.cos2dxxx 解解 原式原式 xxd tandxxxx tantan.coslntancxxx 例例16 求求.sincos3dxxxx 解解 原式原式 )sin121(2xxddxxxx 22csc21csc21.
10、cot21csc212cxxx )(csc212xxd例例17求求.sin2dxxex 解解 原式原式 )21(sin2xexddxxexexx cos21sin2122 )21(cos21sin2122xxexdxedxxexexexxx sin41cos41sin21222所以所以 原式原式cxxex)cossin2(512例例17 求求.sin2dxxex 另解另解 原式原式 )cos(2xdexdxexxexx222coscos 所以所以 原式原式cxxex)cossin2(512 )(sin2cos22xdexexxdxexxexexxx 2224sinsin2cos例例18 求求x
11、dx3sec解解xdxx2secsecxdx3secxxd tansecxxtansecxdxxxtansectanxxtansecxxtansecxxtansecxdxxsec)1(sec2xdx3secxdxsecxdx3sec|tansec|lnxxxdx3secxxtan(sec21Cxx|)tansec|lndxexexx 1)1(1 xxedex 12xexd 12xexdxex 12对对dxex 1令令tex 1则则12 tex)1ln(2tx dtttdx212 例例19 求求解解原式原式dxex 1 tdttt212 dttt 221112dtt )111(22Ctt )ar
12、ctan(2 1(2xeCex )1arctan 原式原式12 xex 1(4xeCex )1arctan1)42(xexCex 1arctan4例例20 求求dxexxxcos1sin1解解原式原式dxexxxx2cos22cos2sin212dxxex2cos22dxxex2tandxxex2tan2tanxdexdxxex2tanxdex2tan2tanxexCxex2tan解法二解法二:原式原式dxexxxcos1sin1dxexxxx2cos1)cos1)(sin1(dxexxxxxx2sincossincossin1dxxex2sinxdxexcotdxxexsindxxxex2s
13、incos)cot(xdex)sin1(xdexdxxexsinxdxexcot)cot(xdex)sin1(xdexdxxexsinxdxexcotxdxexcotxexcotdxxexsinxexsindxxexsinxdxexcotxexcotCxexsin常见不可积的形式常见不可积的形式:dxx 2cosdxx 2sindxex 2dxex 2dxex 1dxxx sin例例21 求求xxdx11解解 原式原式课堂练习：课堂练习：dxxxxxxx)11)(11(11dxxxx211dxxxx)212121(dxxxxx1212dxxxxdxxx211dxxx22)21()21(1dxx
14、x22)21()21(1txsec2121令tdtttttansec21tan2121sec21dttt)secsec(212cttt|)tansec|ln(tan21cxxxxx|)212|ln2(2122例例22 求求dxxxx)1(arctan22解解 原式原式dxxxxxx)1(arctan)1(2222dxxx2arctandxxx21arctan)1(arctanxxdxxd arctanarctandxxxxx2111arctanx2arctan21dxxxxx)1(1222xxarctanx2arctan21dxxxx)11(2xxarctanx2arctan21xxarcta
15、nx2arctan21|ln xcx)1ln(212例例23 求求dxxxx1arctan1解解 令令12 tx则则tdtdx2tdtttt21arctan2原式原式tx1dtttt221arctan2dttt21arctan2tdtarctan2)1ln(2tttarctan2ct 2arctanxln1arctan12xxcx1arctan2例例24 求求dxexxfxxf xx2)()1()(解解 原式原式dxexxfx)(dxexxfxx2)()1()(1xdfexx)1()(xxedxfdxexxfxx2)()1(xxexf)(dxexxfxx2)()1(dxexxfxx2)()1(
16、xxexf)(dxexxfxx2)()1(cxexfx)(例例25 求求dxxex22)1(tan解解 原式原式dxxxex)1tan2(tan22dxxxex)tan2(sec22dxxextan22xdextan2dxxextan22dxexx2tan2xextan2cxextan2习题习题dxxx2sinsinln)1(xxd cotsinlnxxsinlncotxdxxxcossin1cotxxsinlncotdxx2cotxxsinlncotdxx)1(csc2xxsinlncotCxxcot)1sin(lncot xxCx dxxx2sin)2(2dxxx2cos1xdx21xdx
17、xcos2142xxxdsin2142x)sinsin(21dxxxx42xCxxxcos21sin21dxxx21ln)3()1()1(lnxdxdxx21)1(ln1xxcx1)1(ln1xxcxxln1dxeexx23)4(xxdee22)()3(1cex3arctan31dxxx22cos4sin1)5(xdxtantan2122cx2tanarctan21dxeexx221)6(dxeeexxx22)(21)1()1(2xxeedcex)1arctan(dxxxx)ln1(ln1)7()21(ln)21(ln4112xdx)21(ln)21(ln4112xdxcx2121lnarcs
18、incx)1ln2arcsin(解解原式原式令令tx 4则则原式原式dttdx34,4tx dtttt37242dttt4224令令uttan2则则ududt2sec2uduuu24sec2tan4sec24dxxxx432)8(uduuu24sec2tan4sec24duuu4sincos2uud4sinsin2cu3sin13222 tt2uctt332)2(32cxx343)()2(32dxxx113)9(23dxxxxx1133323dxx3dxxx1132223xdxxx12332dxx2112223xdxx113cxx11ln223x|1|ln3xcx 1ln223x|1|ln2x
19、1132xx11x12xdxexxxarctan32)1()10(解解 令令txtan则则tdtdx2sec原式原式tdtettt23secsectantdtetsintdtetsinttdesintetsintdtetcosttdecostetsin)sin(dttettetsintetcos)sin(dttettetsintetcostdtetsintetsintetcos故故tdtetsintet(sin21ct)cos21 xx1t原式原式tdtetsintet(sin21ct)cosxearctan21cxx211dxxxxx)1ln(1)11(22221)1ln(xdxxdxxxx
20、xx)1221(111222)1ln(122xxxdx 1)1ln(122xxxCx)1ln(122xxx例例17 求求.0,)(22 anaxdxInn为自然数为自然数解解 nnaxdxI)(22dxaxxaxan )(1222222dxaxxadxaxann )(1)(1122221222)()(2112222212axdaxxaIann122212)(1)1(1211nnaxdxnaIa)(1)()1(211122122212dxaxaxxanIannn )()1(2111122212 nnnIaxxanIa122212)()1(2)1(21)1(2 nnaxanxIann122212)()1(2)1(21)1(2 nnnaxanxIannI122212)()1(2)1(21)1(2nnnaxanxIannIcaxadxaxI arctan11221caxaxaxaI )(2arctan2122232caxaxaxaxaxaI 222222453)(4)(83arctan83
