数学分析课后习题答案华东师范大学版

1、. 习题P.182 验证下列等式1?Cx)?)df(x?f(x)dx?f(x)?C(f ） （1）2 （?)(xf(x)fC(x)dx?ff(x)?的一个原函数，所以. 证明 （1）因为是?C)?)?f(xdf(xC?du?u. 所以, （2）因为),y(x)(xy?fx2且通过点, 使得在曲线上每一点处的切线斜率为, 2求一曲线),5(2. 2?x?f2(x)C?2xdxf?(x)dx?xf(x)?. 由导数的几何意义, , 知所以解 2C?y?x5?,5)y(22x?所以时，”知，当, 再由条件“曲线通过点, 于是知曲线为221?x?yC2?5?1?C 从而所求曲线为 , , 解得有2x?

2、ysgnx)?|x|(?,验证. 3是在上的一个原函数 222xx?y?yyxy?y?x0?0x0x?x, , , ; 当; 当时证明 当, 时时, 220?xx?20?sgnx(x2)xxsgn?|x0?0?|x?ylim0?lim , 所以的导数为? 2x0?x?0x?0?x?x? 据理说明为什么每一个含有第一类间断点的函数都没有原函数？4?f上的每一点，I3的证明过程可知：在区间I上的导函数，它在解 由P.122推论要么是连续点，要么是第二类间断点，也就是说导函数不可能出现第一类间断点。因此每 一个含有第一类间断点的函数都没有原函数。 5求下列不定积分21421xx? 33?xx1(?x

3、?dxdx)?1?xdxxdxxdxx?3C?33 4223x. . 313141x 22?C|x|?x?)dx?x?)(x?dx?ln(x?222 3x3x11x112dx? ?C?xxdx?C?222 gg2gx2g2xxx22xx2xxx?dx9)?(42?6?(2?3)dx?2?2(2?3)?3)dx xxx92?46C? 9ln4ln6ln3331?C?dx?arcsinxdx 2222x?4?4x12211?x1x1?1?C?(1?arctanx)dx?dx?(1?)dx? 22233x?x3(1?x?)13(122?C?x?tantanxdx?(secx?1)dx?x 12x11

4、?1cos2?C?x)dx?)(1?cos2sinxxdx?dx?(x?sin2 222222xcos2xcossinx?C(cosx?sinx)dx?sinx?cosxdx?dx? xcosx?sinxcos?sinx221sin1coscos2xxx?Cx?tanx?dx?dx?(?)dx?cot 222222xxcossinx?xxsincoscossinx?tt909)(10?t2tt?9)dtdtC10?3?10(?C? 90)lnln(10?91578 ?Cdxx?x?dxxxx?88 1521?xx1?1?x1x?Cx)(?dx?dx(?2arcsin?)?dx xx1?1222

5、x?1xx?11?. . 12?Cxx?dx?(1?sin2x)dx?cos1dx?2sin2xdx(cosx?sinx) 2111?xcos2xdxcos(cos3x?cosx)dx?(sin3x?sinx)?C 32211xx?3x3xx?x?33xx?3xx?C?3e?3e?e(e?e)dx?(e?3e?3ee)dx?e 33 P.188 习题 1应用换元积分法求下列不定积分：11?dx?sin(3x?4)?C?4)cos(3xxcos(3x?4)d(3?4) 331122222xx22x?C?xedx?eed(2x) 4411)dx1d(2x?C?ln|2x?1| 22x?122x?1

6、11nnn?C(1?x)?(1?x)d1?x)?(1?x)dx 1n?11111?xd(?3)dx?)dx?32222x?x1x?3?3x?331 1xC?arcsin3x?arcsin?332?32x2x21232x?2x?3?C?C?2d(2x?23dx?) 2ln22ln233122?1?1? ?Cx)?(8?33x)?(8?x)?C3?8?3xdx?(83x)d(8?222 9333122dx?13?3?1? C5x)?(7?5x)?7?5x)?C(7?(5?(7?x)d333 102553x7?5112222?C?xsinxdxx?cosxdx?sin 22?2d)(x?dx11 4

7、?xcot(2?)C ?42222)?2sin(x)?x(sin2 44. . xdxdxdx 2?C?tan 解法一： xx2x1?cos22coscos2 22xdxcosdxdx?cosx)dx(1? 解法二： 222x1?cosxxsinxsin1?cos1xdsin?C?x?cot?cotx 2xsinxsin 解法一：利用上一题的结果，有?)2?d(xxdx1?C?C?tan(?)?tan(?x) ?221?sinx24)?x1?cos( 21cosxsinx)dxdxddx(1?C?tanx? 解法二： 222x1?sinxcosx1?sinxcoscosx解法三：dxdxdx?

8、 222x1?sin)12?2?(sinx2?cosx2)(tanxcosx2xdtan2?C?2 21xtan2?)?1(tanx2?)xx)d(22?x)dx?sec(?2?cscxdx?sec( 解法一：?C?x|lncscx?tan(cot2?x)|?C?ln|sec(?2x)? 1sinxdcosx1cosx?1?ln?cscxdx?C?dxdx 解法二： 221cosx2sinx?1cos?sinxx?ln|cscx?cotx|?C cscx(cscx?cotx)?dxcscxdx?解法三： xx?cotcsc)x?cotxd(csc?C|?csc?ln|x?cotx?C xcot

9、cscx?xsin解法四：1 2?dx?cscxdxdx xxxx2coscos2sin2sin 2222. . 1xxx?C?ln?|tan|dcot?ln|cot|?C? x222cot 211x?22?d(1?xC)?1?xdx? 222x1?1x?2x11x12?Cdx?dx?arctan? 422224x4)4?(x?xdxdln?Cln|lnx?|? xxlnxln4111?1x5?C?x)?dx?d(1? 253535105)x(1(1?x)?(1?x)31x14?dxdx? 4284?2x(x?)2 442?x?21x11C?ln|ln?|?C 4442?282x22x?dx1

10、1x?C?ln|1?x|?Cln?(?)dx?ln|x|? x1?x1?x(1?x)xcosx?dx?ln|?sinx|?Ccotxdx sinx5422?xdsin1?sinx)coscosxdx?cosxxdx?( (21) 212435?x?sinsinCx?dx)sinx?sinx?sin(1?2?x?sin 35dxd(2x)?ln|csc2x?cot2x|?C 解法一：(22) sinxcosxsin2xdxcosxdxdtanx?C?x|?ln|?tan 解法二： 2sinxcosxtanxsinxcosx22x)?xcosdx(sindx? 解法三： sinxcosxsinxc

11、osxsinxcosx?)dx?ln|sinx|?(ln|cosx|?C cosxsinx. . xxdeedxdxx?C?arctane?(23) x22x?xx1e?ee?1e?2)?8?3xx2?3d(x2?C?x?8?ln|x|?3dx?(24) 228?x?3xx?3x?82231)?1)2?(xx?2(x?dx?dx(25) 33)x?1(x?1)( 32123?C?ln|x?1?|(?)dx? 2231x?x?1)1?1)(x?1)x?2(xdx? (26) 22ax?ttanx?a 则解 令, 2tdtsecadx22?C?|a?t?ln|sec?tant|?C?ln|x?x

12、1tseca22a?xx11xdx?C?d?(27) 21223222221222)(?ax(x)?a)aaa?x(tx?atan 解法, 则2 令2x1sec1tdtdxa?C?costdt?sintC? 23323222aat?asec)a(x222axa?5x?dx (28) 2x1?t?sinx , 则 解 令55ttxsincos225?tt)?dx?cosdt?cossintdt?d(1 tcos2x1? 513121222352 C)?)(1?x?1cos?cost?Ct?(?x)?(1x?t?cos222 5335x?dx (29) 3x?11 56t?xtdxtx?66 解

13、则令, , . . 26423524x?tt1)1?1?(t?1)(t1?tt)?t?dt(?66?dtdtdx?6? 222t1?t1?1t3x?1 3751tt6t?1t264?Cln|?1)?)dt?6(?t?6(?(t)?t?t 21?2t753t?11 x?t6 其中11?x?dx (30) 1x?1?2t1?x?t1?x?tdt?2dx, , 则 , 令解4t124x?1?1t?dt)?4dt?(21(?)2dx?tdt?t(2t?)2tdt? 1t?1?1t?1tt11?x?2C1|?|x?1?ln?x?1?4x?1?4?t1?4t?4ln|t?|?C 11C1|?|x?1?x?

14、4x?1?4ln 2应用分部积分法求下列不定积分：x2?C?xxarcsinxxdx?xarcsinx?1dx?arcsin 2x1?1?dx?xlnx?x?xlnx?x?Clnxdx x222?xdx?sin2xxdsinx?xsinx?xcosxdx 22?cos2xdxcosx?xsinx?x2sinx?x2xdcosx 2sinx?2xcosx?2sinx?C?xlnx?11?1lnx11lnx1?C?dx?dx?lnxd 322322222xxxxxx42222?C?2xx(lnx)?2lnxxdx?(lnx)dxx(lnx)?2ln?x 2x11122?xx?arctanarcta

15、nxdx?xdxdxxarctan 22221?x1111122?(1?)dx?xarctanx?(xxarctan?x?arctanx)?C? 22222x?1112?1)arctanx?x?(x?C 22. . 11?dxln(ln?dx?ln(lnxx)?)dx xlnlnx11?C)?xln(lnx)?x?dx?dx?xln(lnx xlnlnxxxxarcsin222?dx)?)dx?x(arcsin(arcsinxx 2x?122?x(arcsinx)1?2?arcsinxd?x 1222?dxarcsin12?xxx?x(arcsin?x)?21 2x?122C?21?x?arc

16、sinx?x(arcsinx)2x 23?xdxxtanx?tansecxdx?secsecxdtanx?xsec 32?xdxxdx?xdx?secxtan?secx?secxtan?secsecx(sec?x1) 3?|x?sectanxdx?ln|sec?secxtanxx 13?Cx|sec?tanx|)sec?xdx?secxtanx?ln 所以 2x2222?dxxxxx?a?a?dx? 22ax?2a2222?dx?)?(x?a?xx?a 22a?x2a2222?dx?ax?dxx?x?a? 22ax?2222222?)?x?aadx?axln(?x?xx?a? 12222222

17、?C)ln(x?xax?adx?(xx?aa? 所以 2 类似地可得12222222?)?aCxxaxax?dx?(x?aln(? 2 求下列不定积分：3. . 1a?a1a?x)?Cf(x)f(x)f)(xdx?f(x)df( a?1?(x)1f?df(x)?arctanfdx?(x)?C 22)1?f(x)(x1?f?(x)df(x)f?ln|f(x)dx?|?C f(x)f(x)f(x)f(x)f(x)?Cdf(xf)(x)dx?e?ee? 4证明： 1n?1n?xtanI?I?,?n2,3dxI?tanx ，则， 若 2nn?nn?1n?22n?22n?2?dxtan?1)dx?tan

18、xx?Itandxxsecx(sec?x 证nn?2?xdtantanx?I?. 2?nn?2n?1n?2?xdtantantanxx?(ntan?2)xdtanx? 因为，1n?21n?xdtantanx?tanx. 所以 n?11n?1x?tanII. 从而 2?nnn?1mn?0n?m?dxxsinxn)?cosI(m, 若时，则当m?1n?1xmcos?1sinxI(m,n)?I(m?2,n) m?nm?nm?1n?1xcosn?xsin1?I(m,n?2)n,m?2,3,? ， m?nm?n1m?1nmn?1?xd?sincosxdx,I(mn)?cosxsinx 证 n?11m?1

19、n?1m?2n?2?xdxcosxsinx)x?(m?1sin?cos n?11m?1n?1m?2n2?x)cosdxxsincos?xxsin)(x?m?1(cos1 n?11m?1n?1x?(m?1)(I(m?2?cossinx,n)?I(m,n) n?1. . 1?1nm?1xsin?xcosm?n)I2,n)(m,?I(m?所以 ， nm?nm?1m?1n?1sincosxn?x?(m,n)?I(m,n2)I同理可得 nm?nm? P.199 习题 1求下列不定积分：331?1x?x12?dx)dx?dx?1(x?x 1?x1xx?123xxC?1|?x?ln|x 232)x?4x?2

20、21(?Clndx?(?)dx?解法一： 2|?3xx?4x?3|12x?x?7 解法二：371x?212x?dx?dxdx 2222212?12?xx?7x12?7xx?7x271?7x?121d)(x3?)d(x? 17222212x?x?72?)(x 42413x?2C12?|?ln|xln?7?x? 3?22xC?1ABx1? 解 223x1?x?x(1?x)(?1x?)?x1?x12)?C)(1?xx1?A(1?x)?(Bx 去分母得 33C?21A?1x?0?C?x1?A比较上式两端二，于是. ，得再令，得令. 3?1B0?A?B ，从而，因此次幂的系数得 2?1dx1xdx?dx

21、? 233x31?xx?11x?. . 1x?11112?dxdx|?ln|1?x? 22236x11?x?x?x?11112?dx?)?ln|1?x|?ln(1?x?x 226343)?(x?1221?12x1(1?x)C?ln?arctan 26x1?x?33222211)?(xx?1)11?xdx1(1?x?dx?dxdx? 解 4444222x1?1x1?x1?x1111?11)d(x?)d(?x 111122xxxx?dx?dx? 111122222222?x?xx?x? 2222xxxx11)?(x?)dd(x11 xx? 1122222?)?2(x?(x?) xx11x?2x 1

22、1xx?C?lnarctan? 1242222?x?x22?2x?1?12x2x?arctan?ln?C 4822xx?2x?1dx? 22)?1)(x1(x?1ABx?CDx?E?, 解得 解 令 222221?x)?x(?x?1)1x1?(x1)(111B?C?D?AE?, 于是, , 442dx1dx1x?11x?1?dx?dx? 222222414x?)(x?)?x1?x)(1(?1x1. . x1111112?1)?arctanx?x?(arctanx?)?Cln|x?1|?ln( 2244448x?x1?11|x?1|1?x(ln?2arctan?x?)?C 24x?12x?1x?

23、214x?251?dx?dx?dx 22222242(2x?2x?1)x?1)?1)(2x?2(2xx?22?2x?1)2d(2x14x?dx? 其中 22222(2x?2x?1)(2x?2x?1)2x?2x?1141?d(22x?1)dx?dx? 222222?)1x2x(?2x?1)2(2x?1)?11(2x?1?arctan(2x?1)I的递推公式. P.193关于P.186 例9或 参见教材 k2(2x?1)?1于是，有 x?21152x?15?C)?2x?1dx?arctan( 22222241?1?1)(?2x?1)22xx?2x(2x5x?35?arctan(2x?1)?C 22

24、)12x?2(2x?2求下列不定积分 dx? x5?3cosxtan?t 令，则解 2dxdx2dtdt1d(2t)1?C?arctan2t? 22222x2?53cos1?t1?t1?4t?1(2t)5?3 2t?11xarctan(2tan?)?C 22dxdxdxdtanx? 222222)323cos?2sinx2x?sinx(?tanx32(cos)x?tanx. . 3()dtanx3112?C)x?arctan(tan 32662)x(1?tan 2xcos?sinx?xdx1cosx?sinxdxcos?dx? xcosx?1?tanxcosx?sinx2sin)x?cos?1

25、?sinxcosx1d(sinx?)?(1)dx?(?dx xsinx2cosx?sin2cosx?1C?(x?ln|cosxsinx|)? 2xdxcosxdxsin?I?I ，另解：设 21xx?sinxcosx?sincosxx?sincos?Cx?I?dx?I 则， 21xsincosx?)(cosx?sinxcosx?sinxd?C?x|dx?ln?|cosx?II?sin 21x?sincosx?sinxcosx1dx?C|cosx?sin?x|)x?I?(?ln 所以 12x?tan12dx)(xx?12?1?x?xdxdx? 22xx?x?x1?1?dx1(?2x?1)3dx2

26、?dx?1?x?x? 2222x?1x?1x?x 7的结果）其中（利用教材P.185例51152x?11222?x)?1?x?)dx?arcsin?(?1x?xdx?xx?( 4224252)?2x?x(1)?1dx?xd(2?2?1?xx? 22xx1?1x?xdx1?dx2x?arcsin? 1552xx?1?2)?(?x24所以有 . . 2x?dx 2xx?11?512x?11132x22Carcsin2?1?xx?x?arcsin(x?)1?x 22224553x?72x?122C?arcsin?1x?x? 4851)?d(x1dx 22?Cx|?ln|x?x? 2112xx?2?)

27、(x42x11?dx 2x1?x2x1?t1?tdt?4?t?x?dx ，代入原式得，则解 令 222x?1)t(1?t?12222?11?t?4t?11?tx1?4?t?dt?4?tdt?dt?dx? ? 22222222x1?)t)1?t1?(1?(1x?tt)?211111?dt4?dt?4dt?4dt? 2222222t1?t1(1(1?t)?t1?t)(1?t)?11?t1111?C?dt?2ln|?dt? 222tt1?1?t1)1?tt(1?t)(1?22x?1?x?11C?ln| xx 题 练 习 总 求下列不定积分：11151333x?14x?2244? ?Cx?x?)x?d

28、x(x2?xdxx44124412 35134x. . 111222?dx?xarcsinxdx?xarcsinx?xarcsinxdx 222x1?2212tsinx1t1?cos?)2tdt?(t?sindx?costdt? 其中 222cost2x?112)x(arcsinx?x1? 21122?dxxxarcsinxarcsinx?xdx所以 22x1?1122C)x1?x?x?arcsinx(arcsinx? 2211122Cx1?x?x?arcsinx?arcsinx? 424dx? x?1u?xudu2dx? ，则令 解 1dx2udu?C?u|)?ln|1?21(?)du?2(

29、u? u1?1?ux?1C?|1x|)?2(x?ln xsinsinxsinxsinx?de?2xsinx?2esinxdsineesin2xdx?2sinxcosxdx xsinsinsinxsinxxsinx?C(sinxe?e1?dsinx)2(e)?sinxe?2)?C2?(e?sinx? xuuux?C?(x?1e?2()u?eC)?2e?edx(令xu)edu2u 1dxdx11?C?arcsin?d()? xx1121?xx2?1x1?22xxtsecx? ，解法二：令dxsecttant1?dt?t?C?arccos?C xttantsec21?xx. . )sinxd(cos

30、x?1?tanxcosx?sinx?dxdx? xsincosxtanx?cosx?sinx1?C?sinx|?lncosx ?xtan1?C?)|)dx?ln|cos(?xdx?tan(?x 4?tanx4122132)?)2?3(xx?x(x?2?C?ln|x?2|dx?dx 3322?x)x?x?2)2x(?2)(1dxdx322?Cx?x?tandx?(1?tanx)tanx?tan?sec 423xxcoscosxcos21?2422?dx(x)dx?sinxdx?)(sin 2xcos4111?2?dx)2cos2x2x)dx?(1?(1?2cos2x?cos 24411sin4x31xC4?sinx?C?x?sin2xsin?(x?2x?) 32288445x?dx 234?3x?x5?5xx?dxdx? 解 223)2x?1)(xx?3x?4?(C5ABx? 令 2221x?x?)?2(x?2)x(x?1)(2)1?C(x(x?1)(x?2)?x5?A(x?2?B 去分母得：22?B?A1?C? ，解得： 3311215x?2?dx?dx?dxdx? 所以 232213?3xx?)2x3?x4?(x122x?