第3-1不定积分的第一类换元积分法课件.ppt

1、第一节不定积分(之二)第一类换元积分法第一类换元积分法 第一类换元法第一类换元法定理定理.,)(有原函数设uf,)(可导xgu 则有换元公式xxgxgfd)()(uufd)()(xgu)(d)(xgxgf(也称换元法换元法即xxgxgfd)()(,凑微分法凑微分法)cos5xdx132dxx1(cos5)55xdx1ln 322xC1(sin5)5xC例例1.1.求求解解:cos5xdx1(cos5)(5)5x dx例例2.2.求求解解:132dxx112232dxx11(23)232dxx例例3.3.求求21(1)1dxxx x：又如1arcsinCx 2211()dxxx21()11()d

2、xx 原式原式=1(1)dxxx解：解：原式原式=22(1)dxxx221dxx2arctanxC221d1()xaxa例例4.求求.d22xax解解:22dxax)()(112axdaxa1Caxa)arctan(1想到公式21duuCu arctan考虑求84d2xxx222)2(dxxCx)22arctan(21)2(xd例例5.求).0(d22axax解解:2d1()xaxa2d()1()xaxaCax arcsin22dxax试求xxx28d222)1(3dxxCx)31arcsin()1(xd例例6.求.)ln21(dxxxxln21xlnd解解:原式=xln2121)ln21(d

3、xCx ln21ln21例例7.求.e1dxx解法解法1xxe1dxxxxde1e)e1(xdxxe1)e1(dxCx)e1ln(解法解法2 xxe1dxxxde1exxe1)e1(dCx)e1ln()1(elne)e1ln(xxx两法结果一样两法结果一样例例8.求解解:2sin xdx3sin xdx2sin xdx1 cos22xdx12dx1cos2(2)4xdx11sin224xxC2sinsinxxdx2(1 cos)cosx dx 31coscos3xxC 例例9.9.求.dsec6xx解解:原式=xdxx222sec)1(tanxtandxxxtand)1tan2(tan24x5

4、tan51x3tan32xtanC例例10.求.dcossin34xxx解解:原式=xxxxdcoscossin24xdxxsin)sin1(sin24xdxxsin)sin(sin64x551sinx771sinC例例11.11.求求sin4 cos3xxdx解解:原式原式=1sincossin()sin()2利用公式利用公式1(sin7sin)2xx dx1cos714x 1cos2xC例例12.求.dcscxx解：解：xxdcsc2tand2tan1xxCx2tanln2d2cos2sin2cos2xxxxxxxd2cos2sin212tanxxxcotcsc xxsincos12cos

5、2sin2sin2xxx2cos2sinxxCxx|cotcsc|lnxxdsecxxd)2csc(Cxxtansecln同样可同样可求求)2(d)2csc(xxCxx|)2cot()2csc(|lnxxdsec或利用上述结论或利用上述结论Caxaxaln21例例13.求求.d22axx解解:dxaxax)(1dxaxax)11(原式原式=a21axxaxxdda21axax)(d a21ax lnax lnCaxax)(da21有理函数的积分有理函数的积分(拆项法拆项法)一般地，积分一般地，积分)()(1badxbxax)(1bxax)(axAbxB)()()(bxaxaxBbxA)()(b

6、xaxaBbAxBA10aBbABA拆项拆项：abBbaA11)()(1badxbxaxxdbxabxdaxba1111)(a)(bCbxabaxba|ln1|ln1Cbxaxba|ln1 有理函数有理函数:)()()(.1mnxQxP即真分式真分式,)()()1(naxxQ的因式;)()(Q)P(221nnaxAaxAaxAxx可拆分为可拆分为)(,)()()2(2nqpxxxQ的因式.)()(22222211nnnqpxxBxAqpxxBxAqpxxBxA可拆分为可拆分为)04(2 qp)()(xQxPnnnaxaxa110mmmbxbxb110)0,0,(00baNmn例例14.求下列有

7、理函数的积分;651)1(2dxxxx.1)2(3dxxx解解:(1)拆项法6512xxx2xA3xB)3)(2(23)(xxBAxBA1231BABA43BA)3)(2(1xxx原式原式=dxxdxx314213Cxx|3|ln4|2|ln36512xxx23x34xdxxx31)2(拆项法xx 31)1(12xxxA12xCBx)1()()1(22xxxCBxxA)1()(22xxACxxBA100ACBA10,1BCAdxxx31dxxxdxx112|ln x)1(112122xdxCxx)1ln(21|ln2xx 31x112xx假分式相除多项式+真分 式例如例如1123xxx11)1(22xxx112xx2.2.假分式假分式)()()(mnxQxP即dxxxx1123dxxx)11(2dxxxdx112221xCxarctan作业作业练习题练习题3.1（P68）：）：3(6)-(24)xxxd11)132备用题备用题 .求下列积分:)1(d113133xxCx1323xxxxd2132)22xxxd2125)22(x2221)21d(xxxx 52)1(2 x)1d(x2212xx Cx21arcsin5