不定积分换元法例题

不定积分换元法例题不定积分换元法例题不定积分换元法例题【不定积分的第一类换元法】已知求

f(u)duF(u)Cg(x)dxf((x))'(x)dxf((x))d(x)【凑微分】f(u)duF(u)C【做变换，令u(x)，再积分】F((x))C【变量复原，u(x)】【求不定积分g(x)dx的第一换元法的详尽步骤以下：】（1）变换被积函数的积分形式：g(x)dxf((x))'(x)dx（2）凑微分：g(x)dxf((x))'(x)dxf((x))d(x)（3）作变量代换u(x)得：g(x)dxf((x))'(x)dxf((x))d(x)f(u)du（4）利用基本积分公式f(u)duF(u)C求出原函数：g(x)dxf((x))'(x)dxf((x))d(x)f(u)duF(u)C（5）将u(x)代入上面的结果，回到原来的积分变量x得：g(x)dxf((x))'(x)dxf((x))d(x)f(u)duF(u)CF((x))C【注】熟悉上述步骤后，也可以不引入中间变量u(x)，省略(3)(4)步骤，这与复合函数的求导法例近似。__________________________________________________________________________________________【第一换元法例题】1、(5x7)9dx(5x7)9dx(5x7)91d(5x7)1(5x7)9d(5x7)111515(5x7)9d(5x7)(5x7)10C(5x7)10C551050【注】(5x7)'5,d(5x7)5dx,dx1d(5x7)52、lnxdxlnx1dxlnxdlnxxxlnxdlnx1(lnx)2C1(lnx)2C22【注】(lnx)'1,d(lnx)1dx,1dxd(lnx)xxx3（1）tanxdxsinxdxsinxdxdcosxdcosxdcosxcosxcosxcosxcosxln|cosx|Cln|cosx|Ccosx【注】(cosx)'sinx,d(cosx)sinxdx,sinxdxd(cosx)3（2）cotxdxcosxdxcosxdxdsinxdsinxsinxsinxsinxln|sinx|Cln|sinx|Csinx【注】(sinx)'cosx,d(sinx)cosxdx,cosxdxd(sinx)4（1）1dxa1dx1d(ax)axxax1d(ax)ln|ax|Cln|ax|Cax【注】(ax)'1,d(ax)dx,dxd(ax)4（2）1dxx1dx1d(xa)xaaxa1d(xa)ln|xa|Cln|xa|Cxa【注】(xa)'1,d(xa)dx,dxd(xa)4（3）11111111a2dxx2a2dx2axaxadx2axadxxadxx21ln|xa|ln|xa|C1lnxaC2a2axa5（1）secxdxsecx(secxtanx)sec2xsecxtanxdxsecxtanxdxsecxtanxd(tanxsecx)d(tanxsecx)ln|secxtanx|Csecxtanxsecxtanx5（2）secxdx1dxcosxdxcosxdxdsinxcosxcos2xcos2x1sin2xdsinx1111dsinx1lnsinx1C1ln1sinxC1sin2x2sinx1sinx2sinx121sinx6（1）cscxdxcscx(cscxcotx)csc2xcscxcotxdxcscxcotxdxcscxcotxd(cotxcscx)d(cscxcotx)ln|cscxcotx|Ccscxcotxcscxcotx6（2）cscxdxcscx(cscxcotx)csc2xcscxcotxdxcscxcotxdxcscxcotxd(cotxcscx)d(cscxcotx)ln|cscxcotx|Ccscxcotxcscxcotx7（1）1dx1x2dx1x2arcsinxCdxx1dxdxadx7（2）dxaarcsina2a2x2Cx2a1x21x21x2aaaa8（1）1dxx2dx1x2arctanxC11dxdx1dxdx1x8（2）a1aC，（a0）22dxa222a2ax2aarctanaxxa21x1xaa1aa9（1）sin3xcos5xdxsin2xcos5xsinxdxsin2xcos5xdcosx(1cos2x)cos5xdcosx(cos7xcos5x)dcosxcos8xcos6xC869（2）sin3xcos5xdxsin3xcos4xcosxdxsin3xcos4xdsinxsin3x(1sin2x)2dsinx(sin3x2sin5xsin7x)dsinxsin4xsin6xsin8xC43810（1）dx11dx1dlnx1dlnxlnlnxCxlnxlnxxlnxlnx10（2）dx111dlnx1dlnx1Cxln2xln2dxln2xln2xxxlnx11（1）x2xdx22xdxdx22d(x21)arctan(x21)C42x2x42x22x42x21(x21)211（2）xdx12xdx1dx21d(x21)x42x252x42x252x42x2524(x21)22dx2121d(x1)121arctan(x1)C2x22281x2141142212、sinxdxsinx1dx2sinx1dx2sinxdxxx2x2sinxdx2cosxC2cosxC13、e2xdx1e2xd2x1e2xd2x1e2xC22214、sin3xcosxdxsin3xcosxdxsin3xdsinxsin3xdsinxsin4xC415、(2x5)100dx(2x5)100dx(2x5)1001d(2x5)1(2x5)100d(2x5)221(2x5)100d(2x5)11(2x5)101C1(2x5)101C2210120216、xsinx2dxsinx2xdx1sinx2dx21sinx2dx21cosx2C22217、lnxdxlnx1dxlnxdlnx(1lnx)1dlnxx1lnx1lnxx1lnx1lnx1lnxdlnx1dlnx1lnx1lnxd(1lnx)1d(1lnx)1lnx231(1lnx)22(1lnx)2C318、earctanxarctanx1arctanxarctanxarctanx1x2dxe1x2dxedarctanxedarctanxeC19、xdx1xdx1dx21d(1x2)1x21x221x221x21d(1x2)1x2C21x2sinx113120、dxsinxdxdcosxcos2xdcosx2cos2xCcos3xcos3xcos3x21、exxdx1exexdx1exdex1xd(2ex)ln(2ex)C2e222e22、ln2xdxln2x1dxln2xdlnxln2xdlnxln3xCxx323、dxdxd(1x)d(1x)1xx2x)2x)2(2)2x)2arcsinC12x2(12(1(12dxdxd(x1)d(x1)24、271272x2x(x12(x2(x1)2(7)2)4)42222d(x1)2x122x122CC(x1)2(7)2arctanarctan777722225、计算sinxcosxdx，a2b2a2sin2xb2cos2x【解析】因为：(a2sin2xb2cos2x)'a22sinxcosxb22cosx(sinx)2(a2b2)sinxcosx所以：d(a2sin2xb2cos2x)2(a2b2)sinxcosxdxsinxcosxdx1d(a2sin2xb2cos2x)2(a2b2)【解答】sinxcosxdxsinxcosxdx1d(a2sin2xb2cos2x)a2sin2xb2cos2xa2sin2xb2cos2xa2b22a2sin2xb2cos2x1d(a2sin2xb2cos2x)a21a2sin2xb2cos2xCa2b22a2sin2xb2cos2xb2【不定积分的第二类换元法】已知f(t)dtF(t)C求g(x)dxg((t))d(t)g((t))'(t)dtf(t)dtF(t)CF(1(x))C

【做变换，令x(t)，再求微分】【求积分】【变量复原，t1(x)】__________________________________________________________________________________________【第二换元法例题】sin令xtxdx1、xxt2变量复原

sintdt2sint2tdt2sintdttt2costC2cosxCtx2（1）11xdx

令xt1dt212tdt2tdt211dtxt21t1t1t1t变量复原2tln|1t|C2xln|1x|Ctx2（2）11xdx

令1+xtx(t1)2变量复原

1d(t1)212(t1)dt2t1dt211dttttt2tln|t|C21xln|1x|Ct1x341令1t1134xd(t31)4t4(t31)33t2dt3、1xdx34(tt(t32xx(t1)31)1)474347344变量复原(1x)12(t6t3)dt12ttC12(1x)374C414、dxx(1x)

令xt1dt212tdt212dtxt222t(1t)t(1t)1t变量复原2arctantC2arctanxCtx1令ext111dt1115、xdxdlntdtdt1exlnt1t1ttt(1t)t1tt变量复原xln|t|ln|1t|ClnClneCtexx1t1e6、dx令6xt1dt615t2dt6112dt(13x)x62)t32)t36tdt61t21txt(1t(1t变量复原666(tarctant)C66(xarctanx)Ctxmx,nx时，可令kt，其中k为m,n的最小公倍数。【注】被积函数中出现了两个根式xdx3t2t2令x2tdt3tln|1t|C7（1）3xt3231x21t2变量复原32(x2)336x32x2ln|1x2|Ct7（2）11xdxxx

令1xtt2x22dt2t2ln|t1|ln|t1|C2x1t1t21变量复原1x2ln|1x1|ln|1x1|C2txxx1xxnb或naxbt，即可消去根式。【注】被积函数中含有简单根式axcxd时，可令这个简单根式为dx8(1)x8(1x2)

令1t11t81xdtt2dt2dtt6t4t212dtx111111tt111tt8t2t8t2t7t5t3变量复原11111753tarctantCt17x75x53x3xarctanxCx令1t11111lnxxln1ln11lnt2dxttdt8（2）(xlnx)112d12t22dtxt1t11tlnttlntlntt变量复原1x

12(1lnt)dt1d(1tlnt)12C1tlnt1tlnt1tlnt1x1CxC1lnx1lnxx【注】当被积函数中分母的次数较高时，可以试一试倒变换。1sinx9、sinx(1cosx)dx

令tanxt12t2t12t22dt2t1t12d2arctan2x2arctant22t2(11t2)22t(11t2)1t12t1t1t1t1t21dtt2t1ln|t|C2t42变量复原tan2x1ln|tanx|Ct12tanx4222x【注】对三角函数有理式的被积函数，可以用全能公式变换，化为有理分式函数的积分问题。令xasint，|t|10（1）a2x2dx2a2a2sin2tdasinta2cos2tdttarcsinxa令，dxxasint|t|2dasint变量复原xdttCxarcsina2x2xa2a2sin2tCtarcsinatarcsinaaa21cos2tdta2(1cos2t)dta2tsin2tC2222变量复原a2arcsinx1xa2x2Ctarcsinx2a2a令xatant，|t|10（2）dx2datanttarctanxsectdtln|secttant|C22222axaaatant变量复原arctanxa

ln|xa2x2|Cln|xa2x2|Caa因为：所以：

(xa2x2)'2a2x2a2x2a22x2)'dx2a2x2dxa2dx(xaa2x2即：a2x2dx1(xa2x2)'dxa21dx2a2x21xa2x2a2ln|xa2x2|C22令，dxdasect10（3）2sectdtln|secttant|C2a222a2xasect变量复原xasect

ln|xx2a2|Cln|xx2a2|Caa因为：(xx22)'2x22a2aax2a2所以：(xx2a2)'dx2x2a2dxa2dxx2a2即：x2a2dx1(xx2a2)'dxa21dx2x2a21xx2a2a2ln|xx2a2|C22【注】当被积函数中出现a2x2,a2x2,x2a2因子时，可以用三角变换，化为三角函数的积分问题。_______________________________________________________________________________________________【附加】【应用题】已知生产x单位的某种产品，边缘单位成本是C'(x)(C(x))'100个单位时，成本为102，xx2，产量为1又知边缘收益为R'(x