分部积分法

1此时通常考虑应用分部积分法.第五节分部积分法第五章例xxxdcos两个不同类型函数乘积的积分，换元法失效!?2由导数公式vuvuuv)(积分得:xvuxvuuvdd分部积分公式xvuuvxvudd或uvvuvudd1)v容易求得;容易计算.一、分部积分公式3二、例题例求.dcosxxxxdxxcos]sincos[2122xdxxxx积分更难进行，显然，选择不当.vu,解法二xdxxcosxxdsinxdxxxsinsin.cossinCxxx解法一2dcosxx21]cosdcos[2122xxxxuvuv4例xdxxarctan)](arctanarctan[2122xdxxxdxxxxx222121arctan21dxxxx)111(21arctan2122.)arctan(21arctan212Cxxxx2arctan21xdxuv115例xdxxln3)](lnln[4144xdxxx.161ln4144Cxxxdxxxx3441ln41一般的，使用分部积分法时按“指三幂对反”的顺序优先选作去和dx凑成dv.v注：指:指数函数；三:三角函数幂:幂函数；对:对数函数反:反三角函数4lnxdx41dxxxxx141ln41446例xxxdexxxxdee)e(dxx例xxxde2xxxxxde2e2Cxxxxx)ee(2e2)e(d2xxxxxxde2e2.eeCxxx分部积分法可多次使用..e)22(2Cxxx22deexxxx)de(2e2xxexxxx7例xxdlnxxxxdlnln.lnCxxx例xxdarcsinxxxxxd1arcsin2.1arcsin2Cxxxuvxxxdlnxxxxxd1lnxxxxdarcsinarcsin)-d(11121arcsin22xxxx8例xxxxd1arctan221darctanxx原式xxxxxd111arctan1222.)1ln(arctan122Cxxxxxxxxd11arctan122Caxxxax)ln(d12222,1122xxx9例xxxxdsincos3)(sindsin3xxx)sin1(d212xxxxxxdcsc21sin222.)cotcsc(212Cxxx10练习：求下列不定积分xxxd2cos)1(2sind2xxxxxxd2sin22sin2.2cos42sin2Cxxxxxxd)1(ln)2(2xx11dlnxxxxxd)1(11lnxxxxxd)111(1ln.1ln1lnCxxxxxxxd2cos)1(xxxd)1(ln)2(2xxxdcos)3(2.cos2sin)2(2Cxxxxxxxdcos)3(211例分部积分法与换元法结合:令tx12,.de12xx求解，)1(212tx，ttxddtttde原式Ctt)1(e.)112(e12CxxttdettttdeeCtttee12例xxxd1ee2，令tx1e，)1ln(2tx，tttxd12d2解：,1e2txtttdt12t)1(222原式t)dt122（Ctt)3123（Ctt3322Cxx23)1e(321e213例xxxd1ee2tttd1txe令xxxde1eetttd111Ctt12)1(3223.1e2)1e(3223Cxxttttd11d1)1d(11)1d(1tttt14练习:xxxdeearctan.11.解一令xueuuuud1arctan原式)1(darctanuuuuuuud111arctan12,d1d,lnuuxux则uuuuuud]11[arctan12xxdsin.2uuuuuuud1d1arctan12)1(d1121d1arctan122uuuuuu15Cuuuu)1ln(21lnarctan12Cxxxx)e1ln(21earctane2解二彻底换元令xetarctan则txtextanln,tantdttdx2sectan1dxeexxarctantdtttt2sectan1tan原式dttt2sin1ttdcot)1(d1121d1arctan122uuuuuu16ttcotdtdtttcotcotCttt|csc|lncotCeeeexxxx21lnarctanCexeexxx)1ln(21arctan2xxdsin.2tttdsin2ttcosd2ttttdcos2cos2Ctttsin2cos2.sin2cos2Cxxxtdtdxtxtx2,2令textanxetarctan17例xxxdsinexxdesinxxxxxdcosesinexxxxdecossinexxxxxxdsine)cos(sinexxxdsine)cos(sin2exxx.C.C分部积分过程中出现循环，实质上是得到待求积分的代数方程，移项即可求得所求积分。注意最后一定要加上积分常数C.]cosdecose[sinexxxxxx18例dxx3sec解dxx3secxdxx2secsecdxxxxxsectantansec2xdxxdxxx3secsectansecxdxxxxx3sec|tansec|lntansecCxxxxxdx|tansec|ln21tansec21sec3dxxxxxsec)1(sectansec2xxdtansec19例.求.dxI23)1(2x解先换元后分部令,arctanxt即,tantx则teIt3secttdsec2ttetdcostetsinttetdsintetsinttetdcostetcos故CettIt)cos(sin2121xearctantx121x21xx211xCexarctan,dtsectandd2ttx20练习：求下列不定积分xxdlncosxxxxxxd1lnsinlncosxxxxdlnsinlncosxxxxxxxxd1lncoslnsinlncos,dlncoslnsinlncosxxxxxx.)lnsinln(cos2dlncosCxxxxx移项,得21例.已知的一个原函数是求解:xxfxd)()(dxfx)(xfxxxfd)(xxxcosCxxcosxsinCxxcos2说明:此题若先求出再求积分反而复杂.xxfxd)(xxxxxxdcos2sin2cos222*例10已知)(xf的一个原函数2xe,求dxxfx)(.解dxxfx)()(xxdfdxxfxxf)()(222xex.2Cex)(2xexCex223备用题1.xxxxd1ee，令tx1e，)1ln(2tx，tttxd12d2ttttttd12)1ln()1(222ttd)1ln(22tttttd14)1ln(2222Cttttarctan44)1ln(22.1earctan41e)42(Cxxx24作业：P228习题5.51.(2)(4)(6)(8)(10)(12)2.