1不定积分第一类换元法(凑微分法)一、方法简介设)(xf具有原函数)(uF,即)()('ufuF,CuFduuf)()(,如果U是中间变量,)(xu,且设)(x可微,那么根据复合函数微分法,有dxxxfxdF)(')]([)]([从而根据不定积分的定义得)(])([)]([)(')]([xuduufCxFdxxxf.则有定理:设)(uf具有原函数,)(xu可导,则有换元公式)(])([)(')]([xuduufdxxxf由此定理可见,虽然dxxxf)(')]([是一个整体的记号,但如用导数记号dxdy中的dx及dy可看作微分,被积表达式中的dx也可当做变量x的微分来对待,从而微分等式dudxx)('可以方便地应用到被积表达式中。几大类常见的凑微分形式:○1)()(1)(baxdbaxfadxbaxf)0(a;○2xdxfxdxxfsin)(sincos)(sin,xdxfxdxxfcos)(cossin)(cos,xdxfxdxxftan)(tancos)(tan2,xdxfxdxxfcot)(cotsin)(cot2;○3xdxfdxxxfln)(ln1)(ln,xxxxdeefdxeef)()(;○4nnnnxdxfndxxxf)(1)(1)0(n,)1()1()1(2xdxfxdxxf,)()(2)(xdxfxdxxf;○5xdxfxdxxfarcsin)(arcsin1)(arcsin2;2xdxfxdxxfarctan)(arctan1)(arctan2;○6复杂因式【不定积分的第一类换元法】已知()()fuduFuC求()(())'()(())()gxdxfxxdxfxdx【凑微分】()()fuduFuC【做变换,令()ux,再积分】(())FxC【变量还原,()ux】【求不定积分()gxdx的第一换元法的具体步骤如下:】(1)变换被积函数的积分形式:()(())'()dxgxfxxdx(2)凑微分:()(())((')))(()xgxdxdxdxfxfx(3)作变量代换()ux得:()(())'()()()()gxdxfxxxxdxfd()ufud(4)利用基本积分公式()()fuduFuC求出原函数:()(())'()(())()gxdxfxxdxfxdx()()duuCfuF(5)将()ux代入上面的结果,回到原来的积分变量x得:()(())'()(())()gxdxfxxdxfxdx()()fuduFuC(())FxC【注】熟悉上述步骤后,也可以不引入中间变量()ux,省略(3)(4)步骤,这与复合函数的求导法则类似。3二、典型例题○1)()(1)(baxdbaxfadxbaxf)0(a;例1.dxx2010)12(例2.231xx[1]例3.322)1(1xxxdx[1]例4.dxxxx431[1]1.解:令12xu,dxdu2,CxCudxx2011)12(21201121)12(2011201120102.解:令2xt,231xxtdttttdt1)11(21121)1(1121)1(121tdttdtCtt1221)1(322123Cxx22321)1(313.解:322)1(1xxxdx32222)1()1()1(21xxxd令tx21原式ttdttdtttdt1)1(1212123CxCt2112124.解:dxxxx431dxxxdxxx4431142441211)1(41xdxxxd4Cxx24arcsin211241Cxx)1(arcsin2142○2xdxfxdxxfsin)(sincos)(sin,xdxfxdxxfcos)(cossin)(cos,xdxfxdxxftan)(tancos)(tan2,xdxfxdxxfcot)(cotsin)(cot2;例1.dxxtan[2]例2.dxxx2sin[2]例3.dxxxx2sin1cossin1[1]例4.xxdx4cossin[1]例5.xxdx3cossin[1]例6.dxxxxx44cossincossin[1]例7.设ba,为常数,且0a,计算dxxbxaxI2222cossintan[1]1.解:设xucos,xdxdusin,xdxdusindxxtandxxxcossinCxCuudu)ln(cos)ln(2.解:dxxx2sinxdxxxxxdcotcot)(cotCxxxsinlncot3.解:dxxxx2sin1cossin1xxdxxdxdx222sin21)(sincos2)(coscos2)arctan(sincos2cos2ln221)1sec2(cos22xxxxxdxxxdxxx2tan21tan)arctan(sincos2cos2ln221Cxxxx)tan2arctan(21)arctan(sincos2cos2ln2214.解:xxdx4cossindxxxxxdxxxdxxxxx2224422cossincossincossincossincossinxdxxxdxxdsincoscoscoscos24Cxxxxcotcsclncos1cos31355.解:xxdx3cossinxdxxxxxxdxtancostancossincostan2224xdxxtantantan12Cxxtanlntan2126.解:令xu2,再令uvcos,有duuuudxxxxdxxxxx222244sin21cossin412sin212cos2sin21cossincossin222121cos2121coscos41vdvuuudCvarctan21Cx)2arctan(cos217.解:2222222tantantan)tan(costanbxaxxddxbxaxxICbxaabxabxada)tanln(21tan)tan(2122222222222○3xdxfdxxxfln)(ln1)(ln,xxxxdeefdxeef)()(;例1.)ln21(xxdx[3]例2.dxex5[2]例3.dxeexx43[2]例4.xxdx2ln1[2]例5.dxeexx22)1(1[1]例6.dxxxxx4932[1]例7.dxexexx2[1]例8.dxxxxsincostanln[2]1.解:)ln21(xxdxxxdln21lnCxxxdln21ln21ln21)ln21(212.解:令xu5,dxdu56dxex5CeCeduexuu55151513.解:令xeu43,dxedux4,dxeexx43Cuduuln41141Cex)43ln(414.解:令xuln,dxxdu1xxdx2ln1Cuduuarcsin112Cx)arcsin(ln5.解:dxeexx22)1(1dxeeexxx22222)1(2)1(dxeexxx222)1(2222)1()1(4xxeedxCexx2146.解:dxxxxx49321])21[(])23[(23ln11)23()23(22xxxxddxCxx1)23(1)23(ln)2ln3(ln21Cxxxx2323ln)2ln3(ln217.解:dxexexx2)2(22)2(xxxexdeexddxeexxx2222令22tex,22tex,)2ln(2tx,dtttdx222原式dttttexx222222dtttexx222224227dttexx)221(4222Cttexx2arctan218422Ceeexxxx22arctan2424228.解:dxxxxsincostanlnxdxxtantantanln)tan(lntanlnxxdCx2)tan(ln2○4nnnnxdxfndxxxf)(1)(1)0(n,)1()1()1(2xdxfxdxxf,)()(2)(xdxfxdxxf;例1.dxxex3[2]例2.dxxx231[4]例3.dxxxx11[4]例4.)lnln(bxaxxdx[1]例53222)1(1dxxxx[1]例6.)(xaxdx)0(a[1]例7dxxx1arcsin[1]1.解:xdxxd21dxxex3)3(32233xdexdexxCex3322.解:dxxx231)1()111(21121222222xdxxdxxxCxx22321)1(3183.解:dxxxx112222111)1(xdxxxxdxdxxxx对于右端第一个积分,凑微分得)1()1(122122xdxdxxxCx21第二个积分中,用代换txsindxxx221dtttdttt2cos1coscossin22Ctt2sin412Cxxx2121arcsin21原式Cxxx21)2(21arcsin214.解:)lnln(bxaxxdxdxbaxbxax)(lnln)(lnln1)(lnln1bxdbxbaaxdaxbaCbxaxba])(ln)[(ln)(3223235.解:3222)1(1dxxxx)11()11()1()11(3232xdxxdxC35)511(536.解:)(xaxdxCaxxaxdarcsin2)(227.解:dxxx1arcsin)1(arcsin2xdxxdxxxx112arcsin12Cxxx2arcsin12○5xdxfxdxxfarcsin)(arcsin1)(arcsin29xdxfxdxxfarctan)(arctan1)(arctan2;例1.dxxx2arccos2110[3]例2.dxxxx)1(arctan[4]例3.dxxxx)1(arctan1[1]例4.324)(arcsin1xxxdx[1]例5.dxxxxx22211arcsin[1]1.解:dxxx2arccos2110Cxdxx10ln210arccos10arccos2arccos22.解:dxxxx)1(arctan)(arctanarctan21arctan2xdxxdxxCx2)(arctan3.解:dxxxx)1(arctan1dxxxx])(1[arctan12)1(arctanarctan12xd