不定积分换元法积分技巧

第二节不定积分的换元积分法一、第一类换元法二、第二类换元法三、基本积分表⑵问题xdx2cos,2sinCx解决方法利用复合函数，设置中间变量.过程令xt2,21dtdxxdx2cosdttcos21Ctsin21.2sin21Cx一、第一类换元法在一般情况下：设),()(ufuF则.)()(CuFduuf如果)(xu（可微）dxxxfxdF)()]([)]([CxFdxxxf)]([)()]([)(])([xuduuf由此可得换元法定理设)(uf具有原函数)(uF，CxFdxxxf)]([)()]([)(])([xuduuf第一类换元公式（凑微分法）说明使用此公式的关键在于将dxxg)(化为.)()]([dxxxf观察重点不同，所得结论不同.)(xu可导，则有换元公式定理1例1求.2sinxdx解（一）xdx2sin)2(2sin21xxd;2cos21Cx解（二）xdx2sinxdxxcossin2)(sinsin2xxd;sin2Cx解（三）xdx2sinxdxxcossin2)(coscos2xxd.cos2Cx例2(1)求.231dxx解,)23(23121231xxxdxx231dxxx)23(23121xu23令Culn21.23ln21Cxduu121例2(2)求.511dxx解,)51(51151511xxxdxx511dxxx)51(51151duu151Culn51.51ln51Cxdxbaxf)(baxuduufa])([1一般地dxbaxf)(baxuduufa])([1xu51令例3求.12dxex解dxex1212xu令dueu21Ceu21Cex1221)12(2112xdex例4(1)求.12dxxx解dxxx2121xu令duu121Culn21.)1ln(212Cx)21(1122xdx)1(112122xdx例4(2)求.12dxxx解dxxx21)1(112122xdx.Cu21xu令duu21.12Cx例5求.)1(3dxxx解dxxx3)1(dxxx3)1(11)1(])1(1)1(1[32xdxx.)1(21112Cxx例6求解dxxxcossin.tanxdxxdxtan)cos(cos1xdxCxcosln同理可得)(coscos1xdx（使用了三角函数恒等变形）xdxcotCxsinlnxdxtanCxcosln例7求.122dxxa解dxxa221dxaxa222111axdaxa2111.arctan1Caxa例8(1)求.25812dxxx解dxxx25812dxx9)4(12dxx13413122341341312xdx.34arctan31Cx例8(2)求.4412dxxx解dxxx4412dxx2)2(1.21Cx)2()2(12xdx例8(3)求.122dxax解dxax221dxaxaxa1121dxaxdxaxa1121)(1)(121axdaxaxdaxaCaxaxalnln21Caxaxaln21例8(4)求.2312dxxx解dxxx2312dxxx)2)(1(1dxxx1121)1(11)2(21xdxxdxCxx1ln2lnCxx12ln求),(12为常数badxbaxx由例8可知:.42的符号确定可由ba,042badxnmxdxbaxx22)(11,042badxmxdxbaxx22)(11,042badxnxmxdxbaxx))((112例9求.122dxxa解xxad122xaxad11122axaxd112.arcsinCaxdxxa221.arcsinCax例10(1)求.1arctan2dxxx(2)求.11arctandxxxx例10(1)求.1arctan2dxxx解dxxx21arctan)(arctanarctanxdx.)(arctan212Cx(2)求.11arctandxxxx解dxxxx11arctan)(arctanarctan2xdx.)(arctan2Cx例11(1)求.1dxeexx(2)求.11dxex(3)求.)11(12dxexxx(4)求.1212dxexxx例11(1)求.1dxeexx解dxeexx1)1(11xxede.)1ln(Cex例11(2)求.11dxex解dxex11dxeeexxx11dxeexx11dxeedxxx1)1(11xxededx.)1ln(Cexx例11(3)求.)11(12dxexxx解,1112xxxdxexxx12)11()1(1xxdexx.1Cexx(4)求.1212dxexxx解,1122xxxdxexxx2121)1(212xdexCex21例12求.12321dxxx原式dxxxxxxx123212321232dxxdxx12413241)12(1281)32(3281xdxxdx.121213212133Cxxdxxx41232解例13求.)1(100dxxx解dxxx100)1(tx1dttt100)1(dttt)(100101Ctt101102101102Cxx101)1(102)1(101102例14(1)求解.sin12sin2dxxxdxxx2sin12sindxxxx2sin1cossin2.)sin1ln(2Cx)(sinsin1122xdx例14(2)求解.cos11dxxdxxcos11dxxxxcos1cos1cos1dxxx2cos1cos1dxxx2sincos1)(sinsin1sin122xdxdxx.sin1cotCxx例15(1)求解.dcos2xxdxx2cosdxx22cos1.2sin4121Cxx)2()2cos1(41xdx例15(1)求.dcos2xx(2)求.dcos3xx解dxx3cosdxxxcoscos2)(sin)sin1(2xdx.sin31sin3Cxx例15(1)求.dcos2xx(2)求.dcos3xx解(3)求.dcos4xxdxx4cosdxx222cos1.4sin3212sin4183Cxxxdxxx)2cos2cos21(412例16求解.cossin52xdxxxdxx52cossin)(sincossin42xxdx)(sin)sin1(sin222xdxx)(sin)sinsin2(sin642xdxxx.sin71sin52sin31753Cxxx说明当被积函数是三角函数相乘时，拆开奇次项去凑微分.例17求解.2cos3cosxdxx)],cos()[cos(21coscosBABABA),5cos(cos212cos3cosxxxxdxxxxdxx)5cos(cos212cos3cos.5sin101sin21Cxx例18求解（一）dxxsin1.cscxdxxdxcscdxxx2cos2sin2122cos2tan12xdxx2tan2tan1xdxCx2tanln.cotcsclnCxx（使用了三角函数恒等变形）解（二）dxxsin1xdxcscdxxx2sinsin)(coscos112xdxxucosduu211duuu111121Cuu11ln21.cos1cos1ln21Cxx类似地可推出.tanseclnsecCxxxdx例19求解.cossin12dxxxdxxx2cossin1dxxxxx222cossincossindxxxxsin1cossin2dxxxdxcsc)(coscos12.cotcsclnsecCxxx例20求解.sec4xdxxdx4sec（使用了三角函数恒等变形）dxxx22secsec)(tan)1(tan2xdxCxxtantan313例21(1)求.)ln21(1dxxx解dxxx)ln21(1)(lnln211xdx)ln21(ln21121xdx.ln21ln21Cx例21(2)求.)ln(ln12dxxxx解dxxxx2)ln(ln1dxxxxx22)ln1(ln1dxxxxx22ln1)ln1(1)ln()ln1(12xxdxxCxxln11Cxxxln例22求解.2arcsin412dxxxdxxx2arcsin41222arcsin2112xdxx)2(arcsin2arcsin1xdx.2arcsinlnCx例求解.d2sec12xxdxx2sec12xxxdcos21cos2)d(sinsin312xxCx)sin32arcsin(21例求解.d)4(1xxxxxd)(422原积分Cx2arcsin2xxd)2(412原积分Cx22arcsin问题?125dxxx解决方法改变中间变量的设置方法.过程令txsin,costdtdxdxxx251tdtttcossin1)(sin25tdtt25cossin（应用“凑微分”即可求出结果）二、第二类换元法其中)(1x是)(tx的反函数.证设为的原函数,)(t)()]([ttf令)]([)(1xxF则dxdtdtdxF)()()]([ttf,)(1t设)(tx是单调的、可导的函数，)(1)()]([)(xtdtttfdxxf即有换元公式:并且0)(t，又设)()]([ttf具有原函数定理2,)()]([1的原函数是则xfx,)(t第二类积分换元公式CxFdxxf)()(,)]([1Cx)(1)()]([)(xtdtttfdxxf