1、不定积分-PPT课件二、二、基本积分公式基本积分公式(1)dx xCxx d)2(Cx111xxd)3(Cx ln)1(4)e dxx Cex(5)dxax Caaxln。Cxsinxx2cosd)8(xxdsec2Cx tanxxdsin)7(Cx cosxx2sind)9(xxdcsc2Cx cotxxdcos)6(xxxdtansec)10(Cx secxxxdcotcsc)11(Cxcscxxdtanln cos xCxxdcotln sinxC。22dxax22dxxaxxdsecCxxtanseclncsc dx xln csccotxxC1ln2axCaax2d1xxCx arc
2、tan2d1xxCxarcsin22dxax1arctanxCaa22dxaxarcsinxCa22ln xxaC。)(d1)(d1dbxaaxaaxxxd1)(lndxxxd)(d2x21a21)(d2bax)1()(d11d1xxxxxade)e(d1xaa三、三、常见凑微分常见凑微分。xaxdcos)(sind1axaxaxdsin)(cosd1axasec tan dxx x d(sec)xxxd112)(arctandxxxd112)(arcsindx一般地：一般地：xxfd)()(dxfxxdsec2)tan(dx。四、第二类换元法四、第二类换元法令令1.被积函数含被积函数含令令a
3、xbtnaxbndaxbcxndaxbtcxn。2.被积函数含被积函数含22xa 令令taxsin22xa 令令令令taxtantaxsec22ax cbxax2先配方，再作适当变换先配方，再作适当变换(有时用倒代换有时用倒代换1xt简单）。简单）。)()()(xQxPxR nnnaxaxa110mmmbxbxb110五、有理函数真分式的积分五、有理函数真分式的积分:nnnaxaxa110()nm分母在实数范围内因式分解若分母含因式()kxa若分母含既约因式2()kxpxq，则对应的部分因式为122()()kkAAAxaxaxa，则对应的部分因式为11222222()()kkkB xCB xC
4、B xCxpxqxpxqxpxq。ddu vuvvu六.分部积分公式分部积分公式duvu v xxxxande.dsinxxaxn.dcosxxaxnxxxndlnxxxdarctanxxxdarcsinxbxexadsinxbxexadcos注：下列题型用分部积分法；。不定积分（典型例题）（典型例题）例1221(sin)cos2()tanfxxx，求221(sin)cos2()tanfxxx22cos12sin()sinxxx 2221 sin12sinsinxxx 2212sinsinxx1()2fxxx1(2)dxxx2ln|xxC()f x()f x解：一、由 求()f x()fx例2
5、在()f x0,)上定义，在(0,)内可导，()g x在(,)内定义且可导，(0)(0)1fg0 x 时，()f x()32g xx()fx()1g x(2)fx2(2)121gxx 求()f x，()g x的表达式.解:0 x 时，()f x()g x x1C()fx()gx231x()f x()gx32xxC1C 02C 2()f x 21x()g x 1x()gx31xx0 x 时，(2)fx2(2)121gxx 例2在()f x0,)上定义，在(0,)内可导，()g x在(,)内定义且可导，(0)(0)1fg0 x 时，()f x()32g xx()fx()1g x(2)fx2(2)1
6、21gxx 求()f x，()g x的表达式。答案：()21,(0)f xxx31,0()1,0 xxg xxxx例32min,6dxxx分段函数不定积分的求法：(1)各段分别积分，常数用不同 C1，C2 等表示；(2)根据原函数应该在分段点连续确定 C1、C2 的关系，用同一个常数 C 表示。二、分段函数求不定积分：例32min,6dxxx232y x6 y x2min,6xx26,2,236,3 xxxxxx解:2min,6dxxx2 x21162xxC23 x3213xC23162xxC3x2min,6dxxx2 x21162xxC23 x3213xC23162xxC3x在 2 x连续，
7、12 12C283C12223CC在 3x连续，29C39182C32272 CC2321226,2231,2331276,322xxCxxCxxxCx 2min,6dxxx2()CC自学2max,1dxx2221max,11111xxxxxx 解2max,1dxx由 处连续，得：1x 2122311311113xCxxCxxCx 123222,33CCCC例4()f x定义在 R 上，(0)1f1,(0,1(ln),(1,)xfxx x求()f x。1,0()e,0 xxxf xx()fx10 xex0 x()f x1xC0 x2e xC0 x(0)1f11C在 0 x连续 20C解：三、有
8、理函数的积分：例5222d(1)(1)xaxbxxx的结果中，求常数a，b 的值，使不含反正切函数；不含对数函数；仅含有理函数。例5222d(1)(1)xaxbxxx求 a,b,使不含反正切函数；221(1)1xxxdxABExFln|1|Ax1Bx2d1Exxx2d1Fxxln|1|Ax1Bx2ln(1)2ExarctanFxC不含反正切函数0F222d(1)(1)xaxbxxx2xaxbA2(1)(1)xxB2(1)x2(1)Ex x解：例5222d(1)(1)xaxbxxx求 a,b,使不含反正切函数；221(1)1xxxdxABExF222d(1)(1)xaxbxxx不含反正切函数0F
9、2xaxbA2(1)(1)xxB2(1)x2(1)Ex x32()(2)()()A E xA BE xA E xA B0,21,A EA BEA Ea A Bb0,ab 任意例5222d(1)(1)xaxbxxx的结果中，求常数a，b 的值，使不含反正切函数；不含对数函数；仅含有理函数。221(1)1xxxdxABExF222d(1)(1)xaxbxxx 不含对数函数；0,A0,E1b仅含有理函数0,A0,E0,F1b0,a解：四、凑微分法：例6()()dmaxbpxqx求(0)a 原式=()maxb()paxba pqbadx1()dmpaxbxa()pqba()maxbdx221()2mp
10、axba m()pqba11()(1)maxba mC(2,1)mm解：()()dmaxbpxqx1()dmpaxbxa()pqba()maxbdx2 m时，原式=()pqba11a axbC2ln|paxba1 m时，原式=pxa()pqba1ln|axbaC例7sin222esindexxxxsin222esindexxxxsin221 cos2ed2xxxxsin221e2xx1d(sin2)2xxsin221ed(sin22)4xxxxsin221e4xxC 解求例821 lnd(ln)xxxx求21 lnd(ln)xxxxdx1lnx2x2ln(1)xx21 lnxxln xx21l
11、n(1)xxlnd()xx1ln1 Cxx解：例9241d1xxx求241d1xxx22211d1xxxx221d()1xx1xx211d()1()2xxxx22dxax1arctanxCaa11arctan22xxC解1例9241d1xxx求解2241d1xxx2221d(21)(21)xxxxxx22d2121xxxxxAxBCxD222323()()2424AxBCxDxxdx22dxax1arctanxCaa烦！例10（自学）1 lnd(ln)lnxxxxx解1 lnd(ln)lnxxxxx21 lndlnln(1)xxxxxxxlnln(1)xxxxlnd()xx11d(1)tt t
12、1d(1)tttt t 五、分部积分法分部积分法（被积函数是两类不同函数的乘积）例11ed1 exxxx原式=d(e)1 exxx2d(1 e)xx2()1 exx1 e dxx1 e xt21 exx222d1tttt21 exx224d1ttt21 exx221 14d1 ttt解：例12arcsinarccos dxx x2arcsinarccos(arccosarcsin)12xxxxxxx C原式=arcsin arccosxxx x2arccos1xx2arcsin1xxdxarcsin arccos xxx(arccosarcsin)xx2d(1)xarcsin arccosxx
13、x(arccosarcsin)xx21x21x221111xxdx解：例13()df xxln(1)(ln)xfxx，求eln(1e)ln(1e)xxxxC()f xln(1e)exx()df xxln(1e)dexxxeln(1e)dxxxln(1e)d(e)xxeln(1e)xx1d1exxeln(1 e)xx(1 e)ed1 exxxx解：例14lnd,()nx xnZ12lnln(1)lnnnnxxnxxn nxx1(1)(1)2 lnnn nxx 0(1)!nn IClnd nx xlnnxx1lndnnx xnI1nnIlnnxx递推公式lnnxxn1lnnxx2(1)nnIlnn
14、xx n1lnnxx2(1)nn nI0Ix解：六：三角代换 例15222(1)arcsind1xxxxx原式sinxt2sin t cost2(1 sin)ttcostdt2ddsinttt tt21d(cot)2 ttt21cotcot d2 ttt tt21cotln|sin|2 ttttCtsinxtx121 x2211arcsinln|(arcsin)2 xxxxCx解：223d(2)xxx 例16原式2tanxt2(2tan2)t22tan t22secdtt32262tan2sec d8sectt tt242tand4secttt222sincos d4tt t22sin 2 d
15、16t t2 1 cos4d162ttt2tanxtx222x解：七、倒代换：例1782d(1)xxx 1xt分母含x的因子，分母x的最高次幂m与分子x的最高次幂n满足：2mn原式1xt81t21dtt21(1)t82d1 ttt821 1d1 ttt4221(1)(1)d1 tttt解：例182d221xxxx22dxxa22ln xxaC原式1xt1t21dtt2221tt0t2d22ttt2d(1)1 tt2ln1(1)1 ttC211ln11(1)Cxx (0)x解：2d0,221xxxxx211ln11(1)Cxx 八、sincosdmnxx x型（m，n为正负整数）化为 m，n中至
16、少一个奇数：(sin)d(sin)Rxxm，n均为偶数：降次(cos)d(cos)Rxxm，n均为负偶数（负奇数）：化为(tan)d(tan)Rxx(cot)d(cot)Rxx或或sincosdmnxx x化为m，n中至少一个奇数：(sin)d(sin)Rxx(cos)d(cos)Rxx或例194sindcosxxx答案：3111 sinsinsinln321 sinxxxCx4sindcosxxx42sincosdcosxxxx42sind(sin)1 sinxxx42sin1 1d(sin)1 sin xxx解：sincosdmnxx x m，n均为偶数：降次例2024sincosdxx
17、x1311(sin2sin4sin6)164412xxxxC221sin 2 cosd4xx x1 1 cos4 1 cos2d422xxx1(1 cos2cos4cos4 cos2)d16xxxxx原式1cos coscos()cos()2积化和差公式：解：sincosdmnxx xm，n均为负偶数（负奇数）：化为(tan)d(tan)Rxx(cot)d(cot)Rxx或例2124dsin 2 cos 2xxx3111(2tan2tan 2)2tan23xxCx24dsin 2 cos 2xxx242dtan 2 cos 2 cos 2xxxx421sec 2d(tan2)2tan 2xxx
18、2221(1+tan 2)d(tan2)2tan 2xxx解：九、sincosdsincospxqxxaxbx型（a，b，p，q为常数）解题方法：求待定常数A，B，使sincossincospxqxaxbxsincosaxbx()A分母()B分母例224sin7cosd2sin3cosxxxxx2ln|2sin3cos|xxxC原式=2sin3cosxx(2sin3cos)xx2(1)(2sin3cos)xxdx12d(2sin3cos)2sin3cosxxxxdx解：例23sin2d3sin2cos2xxxx（课外练习）十、两项都难积分2ln1dlnxxx例24lnxCx一项用分部积分，产生
19、另一项的相反项2ln1dlnxxx211ddlnlnxxxx1lnxx21lnxdx21dlnxx解：例2522e(tan1)dxxx2etanxxC22e(tan1)dxxx22e(tan2tan1)dxxxx22e(sec2tan)dxxxx222esecd2 etan dxxx xx x22ed(tan)2 etan dxxxx x2etanxx22etan dxx x22 etan dxx x解：例26221e()d1xxxx2e1xCx221e()d1xxxx22212ed(1)xxxxx22212eded1(1)xxxxxxx22212d(e)ed1(1)xxxxxx2222212
20、2ee ded1(1)(1)xxxxxxxxxx解：十一、含抽象函数的积分含抽象函数的积分例273()dx fxx()f x设的原函数是sinxx，求()d f xxsinxxC或()f xsin()xx3()dx fxx3d()xf x32()3()dx f xx f xx2cossinxxxx3x2cossinxxxx23xsin()xxdx()x23x dsin()xx(cossin)x xxx3(2sinxxxsin2xxxd)x解：例2823()()()d()()f xfx fxxfxfx求221()2()fxCfx原式=23()()()dd()()f xfx fxxxfxfxdfx
21、f23d()fffdfxf2211d()2ffdfxf22112ff122 f f21 fdx解：例2823()()()d()()f xfx fxxfxfx求221()2()fxCfx原式=()()d()()()f xf xfxfx另解2()()()1d()()f xf x fxxfxfx22()()()()d()()f xfxf x fxxfxfx化为参数方程十二、例29d3xxy，其中2()xy xy解题思路：把积分中变量 x、y 换为参变量 t2()xy xy把转化为()()xtyt解令：xyt则：2xytd3xxy21tyt321txt32322d()1311tttttt2d1ttt21ln|1|2tC21ln|()1|2xyC例302dxy，其中22()yxyx解令：ytx则：21xyt2dxy21(1)xtt1(1)ytt32ln|yyCxx23dttt2ln|3ttC
