不定積分換元法例題

.__________________________________________________________________________________________【第一換元法例題】1、(5x7)9dx(5x7)9dx(5x7)915d(5x7)15(5x7)9d(5x7)謝謝閱讀15(5x7)9d(5x7)15101(5x7)10C501(5x7)10C感謝閱讀【注】(5x7)'5,d(5x7)5dx,dx15d(5x7)精品文檔放心下載2、lnxxdxlnx1xdxlnxdlnx精品文檔放心下載lnxdlnx12(lnx)2C12(lnx)2C謝謝閱讀【注】(lnx)'1x,d(lnx)1xdx,1xdxd(lnx)感謝閱讀3（1）tanxdxsinxdxsinxdxdcosxdcosxcosxcosxcosxcosxdcosxln|cosx|Cln|cosx|Ccosx謝謝閱讀【注】(cosx)'sinx,d(cosx)sinxdx,sinxdxd(cosx)感謝閱讀3（2）cotxdxcosxdxcosxdxdsinxsinxsinxsinxdsinxln|sinx|Cln|sinx|Csinx精品文檔放心下載【注】(sinx)'cosx,d(sinx)cosxdx,cosxdxd(sinx)謝謝閱讀4（1）1dx1dx1d(ax)axaxax1d(ax)ln|ax|Cln|ax|Cax【注】(ax)'1,d(ax)dx,dxd(ax)謝謝閱讀4（2）

1 dx 1 dx 1 d(x謝謝閱讀xa xa xa1xad(xa)ln|xa|Cln|感謝閱讀

a)xa|

C【注】(xa)'1,d(xa)dx,dxd(xa)精品文檔放心下載.4（3）1dx111111dx1dxdxdxx2a2x2a22axaxaxaxa2a1a|ln|xa|C1lnxaln|x2aC2axa5（1）secxdxsecx(secxtanx)dxsec2xsecxtanxdxsecxtanxsecxtanxd(tanxsecx)d(tanxsecx)ln|secxtanx|C感謝閱讀secxtanxsecxtanx精品文檔放心下載5（2）secxdx1dxcosxdxcosxdxdsinxcosxcos2xcos2x1sin2xdsinx111dsinx1lnsinx11ln1sinx22C21C1sin2xsinx1sinx1sinx1sinx6（1）cscxdxcscx(cscxcotx)dxcsc2xcscxcotxdxcscxcotxcscxcotxd(cotxcscx)d(cscxcotx)ln|cscxcotx|C精品文檔放心下載cscxcotxcscxcotx謝謝閱讀6（2）cscxdxcscx(cscxcotx)dxcsc2xcscxcotxdxcscxcotxcscxcotxd(cotxcscx)d(cscxcotx)ln|cscxcotx|C精品文檔放心下載cscxcotxcscxcotx感謝閱讀7（1）1dxdxarcsinxC1x21x2xx7（2）1dxdxddxaaa2x2dxa2x2x21x21x2arcsinaCa1aaa8（1）1dxdxarctanxC1x21x28（2）1dxdxdxa2x2a2x2x2a21a

xx1d1d1arctanxaaC，（a0）ax2a1x2aa1aa9（1）sin3xcos5xdxsin2xcos5xsinxdxsin2xcos5xdcosx感謝閱讀.(1cos2x)cos5xdcosx(cos7xcos5x)dcosxcos8xcos6xC869（2）sin3xcos5xdxsin3xcos4xcosxdxsin3xcos4xdsinxsin3x(1sin2x)2dsinx(sin3x2sin5xsin7x)dsinxsin4xsin6xsin8xC43810（1）dx11dx1dlnx1dlnxlnlnxCxlnxlnxxlnxlnx10（2）dx11dx1dlnx1dlnx1Cxln2xln2xxln2xln2xlnx11（1）2xdx2xdxdx2d(x21)arctan(x21)Cx42x22x42x22x42x221(x21)211（2）xdx12xdx1dx21d(x21)x42x252x42x252x42x2524(x21)2dx211d(x1)121arctan(x1)C228x21241x2124212212、sinxdxsinx1dx2sinx1dx2sinxdxxx2x2sinxdx2cosxC2cosxC13、e2xdx12e2xd2x12e2xd2x12e2xC感謝閱讀14、

sin3

xcosxdx

sin3

xcosxdx

sin3

xdsinx

sin3

xdsinx

sin44

x

C15、(2x5)100dx(2x5)100dx(2x5)10012d(2x5)12(2x5)100d(2x5)感謝閱讀.12(2x5)100d(2x5)121011(2x5)101C2021(2x5)101C精品文檔放心下載16、xsinx2dxsinx2xdx12sinx2dx212sinx2dx212cosx2C精品文檔放心下載17、lnxdxlnx1dxlnxdlnx(1lnx)1dlnxx1lnx1lnxx1lnx1lnx1lnxdlnx1dlnx1lnx1lnxd(1lnx)1d(1lnx)1lnx23(1lnx)322(1lnx)12C謝謝閱讀18、earctanxdxearctanx1dxearctanxdarctanxearctanxdarctanxearctanxC1x21x219、xdx1xdx1dx21d(1x2)1x21x221x221x21d(1x2)1x2C21x220、sinxdx1sinxdx1dcosxcos23xdcosx2cos12xCcos3xcos3xcos3x21、exdx1exdx1dex1d(2ex)ln(2ex)C2ex2ex2ex2ex22、ln2xdxln2x1dxln2xdlnxln2xdlnxln3xCxx3.23、dxdxd(1x)d(1x)arcsin1xC12xx22(1x)22(1x)2(2)2(1x)221d(x124、dxdxd(x2)72)x2x21)27(x1)21)2(7)2(x2424(x22d(x1x12)2arctan2C2arctan2x1C(x1)2(7)2777722225、計(jì)算sinxcosxdx，a2b2a2sin2xb2cos2x【分析】因?yàn)椋?a2sin2xb2cos2x)'a22sinxcosxb22cosx(sinx)2(a2b2)sinxcosx謝謝閱讀所以：d(a2sin2xb2cos2x)2(a2b2)sinxcosxdx1x)sinxcosxdxd(a2sin2xb2cos22(a2b2)【解答】sinxcosxdxsinxcosxdx1d(a2sin2xb2cos2x)a2sin2xb2cos2xa2sin2xb2cos2xa2b22a2sin2xb2cos2x1d(a2sin2xb2cos2x)1a2sin2xb2cos2xCa2b22a2sin2xb2cos2xa2b2.【不定積分的第二類換元法】已知f(t)dtF(t)C求g(x)dxg((t))d(t)g((t))'(t)dt【做變換，令x(t)，再求微分】f(t)dtF(t)C【求積分】F((x))C【變量還原，t(x)】11__________________________________________________________________________________________【第二換元法例題】sinx令xtsintdt2sint2tdt2sintdt1、xdxttxt22cost變量還原xCC2costx1令xt1dt212tdt2t1dxdt21dt2（1）1xxt21t1t1t1t2tln|1變量還原t|C2xln|1x|Ctx1令1+xt112(t1)dt2t11dxd(t1)2dt21dt2（2）1xx(t1)2ttx|Ctt變量還原xln|12tln|t|C21t1x131td(t31)41令14xt3、x314xdx(t31)4(t31)2t4(t31)33t2dtx(t31)4.t7t4變量還原3(14x)73(14x)4(t63C7434x74t14、1令xt1dt212tdt21dtdxx(1x)xt2t(1t2)t(1t2)1t22arctant變量還原xCC2arctantx5、1令ext1dlnt11dt1dt11dxdt1exxlnt1t1ttt(1t)t1tln|t|ln|1t|ClntC變量還原exCln1ttex1ex6、dx令6xt1dt616t5dt6t2dt611dt(13x)xxt6(1t2)t3(1t2)t31t21t2變量還原xarctan6x)C6(tarctant)C6(6t6x【注】被積函數(shù)中出現(xiàn)了兩個(gè)根式mx,nx時(shí)，可令kxt，其中k為m,n的最小公倍數(shù)。dx令3x2tt2t2tln|131tdt32t|C7（1）13x2xt32變量還原3(x2)233x2ln|13x2|C2t6x1x11x令xtt2dt2t2ln|t1|ln|t21|C7（2）dx2xxx1t212t1變量還原1x1x1x2x2ln|x1|ln|x1|Ct1xx.naxbnaxb【注】被積函數(shù)中含有簡(jiǎn)單根式或cxd

時(shí)，可令這個(gè)簡(jiǎn)單根式為t，即可消去根式。dx令xt11t81tt18(1)x8(1x2)11t2dtt6t4t21dt11111t2xtt811t2t8t2t7t5t3變量還原11111tarctantCarctanC753t17x75x53x3xxx1lnx令xt11111lnttt18（2）(xlnx)dxd11dt1tlntdt2222xtlnln2tttt1(1lnt)dt111tlnt1tlntd(1tlnt)C1tlnt22變量還原1CxC1ln1t11xlnxxxx【注】當(dāng)被積函數(shù)中分母的次數(shù)較高時(shí)，可以試一試倒變換。精品文檔放心下載1sinx令tanxt12td2arctant12t2dt9、dx21t21t2sinx(1cosx)x2arctant2t1t222t1t21t21t2(11t2)1t2(11t2)11t21ln|t|Ct2dtt22t4tan2xx1x變量還原2tanln|tan|Ct14222x【注】對(duì)三角函數(shù)有理式的被積函數(shù)，可以用萬能公式變換，化為有理分式函數(shù)的積分問題。精品文檔放心下載.令xasint，|t|10（1）a2x2dx2a2a2sin2tdasinta2cos2tdttarcsinxadasintdx令xasint，|t|變量還原xdttCarcsinC2a2x2xa2a2sin2txatarcsinatarcsinaa21cos2ta2a2sin2tdt(1cos2t)dttC2222變量還原a2arcsinx1xa2x2Ctarcsinx2a2adx令xatant，|t|datant10（2）sectdtln|secttant|C2a2x2xa2a2tan2ttarctana變量還原xa2x2|Cln|xa2x2|Cln|aatarctanxa因?yàn)椋?xa2x2)'2a2a2x2a2x2所以：(xa2x2)'dx2a2x2dxa2dxa2x2a2x2dx1(xa2x2)'dxa21dx即：2a2x21a2ln|xa2x2|C2xa2x22dx令xasect，0tdasectsectdtln|secttant|C10（3）2x2a2