高等数学第四章第二节《换元积分法》课件.ppt

1、二、第二类换元法二、第二类换元法第二节一、第一类换元法一、第一类换元法换元积分法 第四四章 第二类换元法第二类换元法第一类换元法第一类换元法xxxfd)()(uufd)(基本思路基本思路 设,)()(ufuF)(xu可导,xxxfd)()(CxF)()(d)(xuuuf)()(xuCuF)(dxFxxxfd)()(则有一、第一类换元法一、第一类换元法定理定理1.,)(有原函数设uf,)(可导xu则有换元公式xxxfd)()(uufd)()(xu)(d)(xxf(也称配元法配元法即xxxfd)()(,凑微分法凑微分法)例例1.求).1(d)(mxbxam解解:令,bxau则,ddxau 故原式原

2、式=muuad1a1Cumm1111)()1(1mbxamaC注注:当1m时bxaxdCbxaaln122)(1d1axxa例例2.求.d22xax解解:22dxax,axu 令则xaud1d21uuda1Cuaarctan1Caxa)arctan(1想到公式21duuCu arctan)(ax例例3.求).0(d22axax21duu想到Cu arcsin解解:2)(1daxax)(d)(xxf(直接配元)xxxfd)()(2)(1)(daxaxCax arcsin22dxax例例4.求.dtanxx解解:xxxdcossinxxcoscosdCx cosln?dcotxxxxxsindco

3、sCx sinlnxxsinsindxxdtan类似Caxaxaln21例例5.求.d22axx解解:221ax)(axax)()(axaxa21)11(21axaxa 原式原式=a21axxaxxdda21axax)(da21ax lnax lnCaxax)(d常用的几种配元形式常用的几种配元形式:xbxafd)()1()(bxaf)(dbxa a1xxxfnnd)()2(1)(nxfnxdn1xxxfnd1)()3()(nxfnxdn1nx1万能凑幂法xxxfdcos)(sin)4()(sin xfxsindxxxfdsin)(cos)5()(cosxfxcosdxxxfdsec)(tan

4、)6(2)(tan xfxtandxeefxxd)()7()(xefxedxxxfd1)(ln)8()(ln xfxlnd例例6.求.)ln21(dxxxxln21xlnd解解:原式=xln2121)ln21(dxCx ln21ln21例例7.求.d3xxex解解:原式=xexd23)3d(323xexCex332例例8.求.dsec6xx解解:原式=xdxx222sec)1(tanxtandxxxtand)1tan2(tan24x5tan51x3tan32xtanC例例9.求.1dxex解法解法1xex1dxeeexxxd1)1(xdxxee1)1(dxCex)1ln(解法解法2 xex1d

5、xeexxd1xxee1)1(dCex)1ln()1(ln)1ln(xxxeee两法结果一样xxsin11sin1121例例10.求.dsecxx解法解法1 xxdsecxxxdcoscos2xx2sin1sindxsindxsin1ln21Cxsin1lnCxxsin1sin1ln21xxtansec解法解法 2 xxdsecxxdsecxxtansec)tan(secxxxxxxxxdtansectansecsec2)tan(secdxx Cxxtansecln同样可证xxdcscCxxcotcscln或xxdcscCx2tanln例例11.求.dsin3 xx例例12.求.dcossin

6、52 xxx例例13.求.dcos2 xx例例14.求.dsec35 xxxtg222d)(2123xax例例15.求.d)(23223xaxx解解:原式=23)(22ax22dxx21222)(aax21)(2122ax)(d22ax 23)(2222axa)(d22ax 22ax 222axaC)2cos2cos21(241xx 例例16.求.dcos4xx解解:224)(coscosxx 2)22cos1(x)2cos21(24cos141xx)4cos2cos2(212341xxxxdcos4xxxd)4cos2cos2(21234141xd23)2d(2cosxx)4(d4cos81

7、xxx83x2sin41x4sin321C例例17.求.d3cos2cosxxx xxexex111xexexxxdd xexxd)1(例例18.求.d)1(1xexxxx解解:原式=xexxxxd)1()1(xexe)1(1xxxexe)(d)111(xxxexexex)1(1xxxxxexexexe)(dxxexexlnxex1lnCCexxxx1lnln分析分析:例例19.求.d)()()()()(32xxfxfxfxfxf 解解:原式原式)()(xfxfxxfxfxfxfxfd)()()(1)()(2 xxfxfxfxfd)()()()(22 Cxfxf2)()(21)()(d(xfx

8、f)()(xfxf小结小结常用简化技巧:(1)分项积分:(2)降低幂次:(3)统一函数:利用三角公式;配方法(4)巧妙换元或配元等xx22cossin1;)2cos1(sin212xx;)2cos1(cos212xx万能凑幂法xxxfnnd)(1nnnxxfd)(1xxxfnd1)(nxnnxxfnd)(11利用积化和差;分式分项;利用倍角公式,如思考与练习思考与练习1.下列各题求积方法有何不同?xx4d)1(24d)2(xxxxxd4)3(2xxxd4)4(2224d)5(xx24d)6(xxxxx4)4(d22221)(1)d(xx22214)4(dxxxxd441241xx2121xd2

9、)2(4x)2(dx2.求.)1(d10 xxx二、第二类换元法二、第二类换元法第一类换元法解决的问题难求易求xxxfd)()(uufd)()(xu若所求积分xxxfd)()(易求,则得第二类换元积分法.难求，uufd)(CxF)()()()(ttft定理定理2.设)(tx是单调可导函数,且,0)(t)()(ttf具有原函数,)(1d)()(d)(xttttfxxf.)()(1的反函数是其中txxt证证:的原函数为设)()(ttf,)(t令)()(1xxF则)(xFtddxtdd)()(ttf)(1t)(xfxxfd)(Cx)(1Ct)(1xt)(1d)()(xttttf则有换元公式例例20.

10、求.)0(d22axxa解解:令,),(,sin22ttax则taaxa22222sintacosttaxdcosd 原式tacosttadcosttadcos22Ca242sin2ttax22xa taxarcsinCxax222122atttcossin22sin2axaxa22例例21.求.)0(d22aaxx解解:令,),(,tan22ttax则22222tanataaxtasecttaxdsecd2 原式 ta2sectasectdttdsec1tanseclnCttax22ax tln22ax a)ln(1aCCCaxx22lnxa1C例例22.求.)0(d22aaxx解解:,时当

11、ax 令,),0(,sec2ttax则22222secataaxtatanxdtttadtansec 原式td ttatansectatanttdsec1tanseclnCttax22ax t1 lnCCaxx22ln)ln(1aCC22ax axa,时当ax令,ux,au 则于是22daxx22dauuCaxx22ln22daxx，时ax 122lnCauu122lnCaxx1222lnCaxxa)ln2(1aCCCaxx22ln说明说明:被积函数含有22ax 时,除采用1shch22tt采用双曲代换taxsh消去根式,所得结果一致.(参考书上 P201-P202)taxch或22ax 或三

12、角代换外,还可利用公式原式21)1(22ta221a例例23.求.d422xxxa解解:令,1tx 则txtdd21原式ttd12tttad)1(2122,0时当x42112tta Cata2223)1(23当 x 0 时,类似可得同样结果.Cxaxa32223)(23)1(d22ta小结小结:1.第二类换元法常见类型第二类换元法常见类型:,d),()1(xbaxxfn令nbxat,d),()2(xxfndxcbxa令ndxcbxat,d),()3(22xxaxf令taxsin或taxcos,d),()4(22xxaxf令taxtan,d),()5(22xaxxf令taxsec第四节讲xxdt

13、an)16(xxdcot)17(xxdsec)18(xxdcsc)19(Cx coslnCx sinlnCxx tanseclnCxxcotcscln2.常用基本积分公式的补充(P203)(7)分母中因子次数较高时,可试用倒代换倒代换,d)()6(xafx令xat xxad1)20(22xxad1)22(22xaxd1)23(22xaxd1)21(22Caxaarctan1Caxaxaln21CaxarcsinCaxx)ln(22xaxd1)24(22Caxx22ln.32d2 xxx解解:原式xxd2)1(122)2()1(dx21arctan21xC(P203 公式(20)例例24.求例例

14、25.求.94d2xxI解解:223)2()2(d21xxICxx942ln212(P203 公式(23)例例26.求.1d2xxx解解:原式=22)()()(d21x(P203 公式(22)2521xCx512arcsin例例27.求.d222 axxx解解:令,1tx 得原式ttatd1221)1(d2122222tataaCtaa11222Cxaax222思考与练习思考与练习1.已知,1d)(25Cxxxfx求.d)(xxf解解:两边求导,得)(5xfx,12xx则1dd)(24xxxxxf)1(xt 令231dttt222d121ttt1(1)1(d)1(212221tt)1(d)1(212221tt23)1(312tCt21)1(2(代回原变量代回原变量)xxxd11)1322.求下列积分:)1(d113133xxCx1323xxxxd2132)22xxxd2125)22(x2221)21d(xxxx 52)1(2 x)1d(x2212xx Cx21arcsin5