第12讲 第一换元积分法

主讲：王淑媛第十一讲4.3两种积分法问题：？4.3.1换元积分法Cxxdxsincos我们知道xdx2cos,2sinCxxdx2cosdxx22cos21dxxx)'2(2cos21xxd22cos21xu2uducos21Cx2sin211.第一类换元法(凑微分法)Cusin21dxxx2sin2求例dxxx)'(sin22原式解22sindxxuduxusin2Cucos.cos2Cx以上两例有一个共同的特点,即dxxxgdxxf)(')]([)(φφduugxu)()(φCuG)(.)]([CxGφ此积分方法称为第一换元积分法,也称凑微分法.)()]([xdxgφφ例1求.dxx231解dxx231dxxx)(2323121duu121Culn21.23ln21Cxxu23)23(23121xdxCxdxxln1例2求解xdxtandxxxxdxcossintandxxxcos)'(cos.coslnCx)(coscos1xdx例3求解dxxxlndxxx)'(lnln原式)(lnlnxxd.)(ln212Cx例4求解dxax221Cxdxxarctan112dxaxa1)(1122原式1)()(12axaxda.arctan1CaxaCaxadxaxarctan1122例5求解1362xxdx4)3()3(2xxd原式.2)3(arctan21Cx例6求解dxax221dxaxaxa)11(21原式)](1)(1[21axdaxaxdaxaCaxaxa]ln[ln21.ln21CaxaxaCaxaxadxaxln21122例7求解dxxa221Cxdxxarcsin1122)(1)(axaxd原式.)arcsin(CaxCaxdxxa)arcsin(122dxxx29621例8求解dxx2)13(31原式22)13()3()13(31xxd.313arcsin31Cx解dxxsin1xdxcsc)(coscos112xdx)(cos1cos12xdxCxx1cos1cosln21类似地可推出例9求.cscxdxdxxx2sinsin.tanseclnsecCxxxdx.cotcsclnCxxCaxaxadxaxln21122Cxx21221cos)1(coslnCxxsincos1lnCxxxdxcotcscln.csc解例10求.secxdxxxdxxxxtansec)tan(secsec原式dxxxxxxtansectansecsec2xxxxdtansec)tan(sec.tanseclnCxxCaxadxaxarctan1122Caxaxadxaxln21122Caxdxxa)arcsin(122.)tanln(secsecCxxxdxCxxxdxcotcscln.csc例11求解.cossin52xdxxxdxx52cossin)(sin)sin1(sin222xdxx)(sin)sinsin2(sin642xdxxx.sin71sin52sin31753Cxxx当被积函数是三角函数相乘时，拆开奇次项去凑微分.)(sincossinxxdx42例12求dxx2cosdxx22cos1Cxx42sin2dxx2cos解))2(221(21xxdcoxdx正弦余弦三角函数积分偶次幂降幂练习:dxx3sin.1求dxx321.2求.)ln51(1.3dxxx求6.求.11dxex5.求.cos11dxxdxxx5.432求习题解答:解Cxxxdxxdxxdxx)cos(coscos)cos(sinsinsin3223311dxx3sin.1求dxx321.2求解)32(32121xdx原式.32ln21Cx.)ln51(1.3dxxx求解dxxx)ln51(1)(lnln511xdx)ln51(ln51151xdxCxln51ln51dxxx5.432求解)5()5(313213xdx原式.)5(92233Cx5.求解.cos11dxxdxxcos11dxxx2cos1cos1dxxx2sincos1)(sinsin1sin122xdxdxx.sin1cotCxx6.求.11dxex解dxex11dxeeexxx11dxeexx11dxeedxxx1)1(11xxededx.)1ln(Cexx