高等数学上册第六版课后习题详细图文答案第四章23377.pdf

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1、高等数学上册第六版课后习题详细答案第四章 习题 41 1.求下列不定积分:(1)dxx21;解 CxCxdxxdxx112111222.(2)dxxx;解 CxxCxdxxdxxx212323521231.(3)dxx1;解 CxCxdxxdxx21211112121.(4)dxxx32;解 CxxCxdxxdxxx3313737321031371.(5)dxxx21;解 CxxCxdxxdxxx12312511125252.(6)dxxmn;解 CxmnmCxmndxxdxxmnmmnmnmn111.(7)dxx35;解 Cxdxxdxx4334555.(8)dxxx)23(2;解 Cxxx

2、dxdxxdxxdxxx2233123)23(2322.(9)ghdh2(g 是常数);解 CghChgdhhgghdh22212122121.(10)dxx2)2(;解 Cxxxdxdxxdxxdxxxdxx423144)44()2(23222.(11)dxx22)1(;解 Cxxxdxdxxdxxdxxxdxx3524242232512)12()1(.(12)dxxx)1)(1(3;解 dxdxxdxxdxxdxxxxdxxx23212323)1()1)(1(Cxxxx25233523231.(13)dxxx2)1(;解 Cxxxdxxxxdxxxxdxxx2523212321212252

3、342)2(21)1(.(14)dxxxx1133224;解 Cxxdxxxdxxxxarctan)113(1133322224.(15)dxxx221;解 Cxxdxxdxxxdxxxarctan)111(111122222.(16)dxxex)32(;解 Cxedxxdxedxxexxx|ln32132)32(.(17)dxxx)1213(22;解 Cxxdxxdxxdxxxarcsin2arctan3112113)1213(2222.(18)dxxeexx)1(;解 Cxedxxedxxeexxxx21212)()1(.(19)dxexx3;解 CeCeedxedxexxxxxx13ln

4、3)3ln()3()3(3.(20)dxxxx32532;解 CxCxdxdxxxxxxx)32(3ln2ln5232ln)32(52)32(5232532.(21)dxxxx)tan(secsec;解 Cxxdxxxxdxxxxsectan)tansec(sec)tan(secsec2.(22)dxx2cos2;解 Cxxdxxdxxdxx)sin(21)cos1(212cos12cos2.(23)dxx2cos11;解 Cxdxxdxxtan21cos212cos112.(24)dxxxxsincos2cos;解 Cxxdxxxdxxxxxdxxxxcossin)sin(cossincos

5、sincossincos2cos22.(25)dxxxx22sincos2cos;解 Cxxdxxxdxxxxxdxxxxtancot)cos1sin1(sincossincossincos2cos22222222.(26)dxxxx)11(2;解 dxxxx211Cxxdxxx41474543474)(.2.一曲线通过点(e2,3),且在任一点处的切线的斜率等于该点横坐标的倒数,求该曲线的方程.解 设该曲线的方程为 yf(x),则由题意得 xxfy1)(,所以 Cxdxxy|ln1.又因为曲线通过点(e2,3),所以有321 3f(e 2)ln|e 2|C2C,C321.于是所求曲线的方程为

6、 yln|x|1.3.一物体由静止开始运动,经 t 秒后的速度是 3t2(m/s),问 (1)在 3 秒后物体离开出发点的距离是多少？(2)物体走完 360m 需要多少时间？解 设位移函数为 ss(t),则 sv3 t2,Ctdtts323.因为当 t0 时,s0,所以 C0.因此位移函数为 st 3.(1)在 3 秒后物体离开出发点的距离是 ss(3)3327.(2)由 t 3360,得物体走完 360m 所需的时间11.73603ts.4.证明函数xe221,exshx 和 exchx 都是xxexshch 的原函数.证明 xxxxxxxxxeeeeeeeexxe222shch.因为xxe

7、e22)21(,所以xe221是xxexshch 的原函数.因为 (exshx)exshxexchxex(shxchx)xxxxxxeeeeee2)22(,所以exshx 是xxexshch 的原函数.因为 (exchx)exchxexshxex(chxshx)xxxxxxeeeeee2)22(,所以exchx 是xxexshch 的原函数.习题 42 1.在下列各式等号右端的空白处填入适当的系数 使等式成立(例如)74(41xddx:(1)dxd(ax);解 dxa1 d(ax).(2)dx d(7x3);解 dx 71 d(7x3).(3)xdx d(x2);解 xdx 21 d(x2).

8、(4)xdx d(5x2);解 xdx 101 d(5x2).(5)1(2xdxdx;解)1(21 2xdxdx.(6)x3dx d(3x42);解 x3dx 121 d(3x42).(7)e 2x dx d(e2x);解 e 2x dx 21 d(e2x).(8)1(22xxeddxe;解)1(2 22xxeddxe.(9)23(cos 23sinxdxdx;解)23(cos 32 23sinxdxdx.(10)|)|ln5(xdxdx;解|)|ln5(51 xdxdx.(11)|)|ln53(xdxdx;解|)|ln53(51 xdxdx.(12)3(arctan 912xdxdx;解)3

9、(arctan 31 912xdxdx.(13)arctan1(12xdxdx;解)arctan1()1(12xdxdx.(14)1(122xdxxdx.解)1()1(122xdxxdx.2.求下列不定积分(其中 a,b,均为常数):(1)dtet 5;解 Cexdedtexxt55551551.(2)dxx3)23(;解 Cxxdxdxx433)23(81)23()23(21)23(.(3)dxx211;解 Cxxdxdxx|21|ln21)21(21121211.(4)332xdx;解 CxCxxdxxdx3232313)32(21)32(2331)32()32(3132.(5)dxeax

10、bx)(sin;解 Cbeaxabxdebaxdaxadxeaxbxbxbxcos1)()(sin1)(sin.(6)dtttsin;解 Cttdtdtttcos2sin2sin.(7)xdxx210sectan;解 xdxx210sectanCxxxd1110tan111tantan.(8)xxxdxlnlnln;解 Cxxdxxdxxxxxdx|lnln|lnlnlnlnln1lnlnlnln1lnlnln.(9)dxxxx2211tan;解 dxxxx2211tan2222211cos1sin11tanxdxxxdx Cxxdx|1cos|ln1cos1cos1222.(10)xxdxc

11、ossin;解 Cxxdxdxxxxxdx|tan|lntantan1tanseccossin2.(11)dxeexx1;解 dxeexx1Cedeedxeexxxxxarctan11122.(12)dxxex2;解.21)(212222Cexdedxxexxx (13)dxxx)cos(2;解 Cxxdxdxxx)sin(21)()cos(21)cos(2222.(14)dxxx232;解 CxCxxdxdxxx2212221223231)32(31)32()32(6132.(15)dxxx4313;解 Cxxdxdxxx|1|ln43)1(11431344443.(16)dttt)sin(

12、cos2;解 Cttdtdttt)(cos31)cos()(cos1)sin()(cos322.(17)dxxx3cossin;解 CxCxxxddxxx2233sec21cos21coscoscossin.(18)dxxxxx3cossincossin;解)sincos(cossin1cossincossin33xxdxxdxxxxx Cxxxxdxx3231)cos(sin23)cos(sin)cos(sin.(19)dxxx2491;解 dxxxdxxdxxx22249491491 )49(49181)32()32(1121222xdxxdxCxx2494132arcsin21.(20)

13、dxxx239;解 Cxxxdxxdxxdxxx)9ln(921)()991(21)(9219222222223.(21)dxx1212;解 dxxxdxxxdxx)121121(21)12)(12(11212 )12(121221)12(121221xdxxdx CxxCxx|1212|ln221|12|ln221|12|ln221.(22)dxxx)2)(1(1;解 CxxCxxdxxxdxxx|12|ln31|1|ln|2|(ln31)1121(31)2)(1(1.(23)xdx3cos;解 Cxxxdxxdxxdx3223sin31sinsin)sin1(sincoscos.(24)d

14、tt)(cos2;解 Cttdttdtt)(2sin4121)(2cos1 21)(cos2.(25)xdxx3cos2sin;解 xdxx3cos2sinCxxdxxxcos215cos101)sin5(sin21.(26)dxxx2coscos;解 Cxxdxxxdxxx21sin23sin31)21cos23(cos212coscos.(27)xdxx7sin5sin;解 Cxxdxxxxdxx2sin4112sin241)2cos12(cos217sin5sin.(28)xdxxsectan3;解 xdxxdxxxxdxxsectantansectansectan223 Cxxxdxs

15、ecsec31sec)1(sec32.(29)dxxx2arccos2110;解 Cxdxddxxxxxx10ln210)arccos2(1021arccos10110arccos2arccos2arccos22arccos2.(30)dxxxx)1(arctan;解 Cxxdxxdxxdxxxx2)(arctanarctanarctan2)1(arctan2)1(arctan.(31)221)(arcsinxxdx;解 Cxxdxxxdxarcsin1arcsin)(arcsin11)(arcsin222.(32)dxxxx2)ln(ln1;解 Cxxxxdxxdxxxxln1)ln()ln

16、(1)ln(ln122.(33)dxxxxsincostanln;解 xdxxxdxxxdxxxxtantantanlnsectantanlnsincostanln2 Cxxdx2)tan(ln21tanlntanln.(34)dxxax222(a0);解 dttadttatdtatatataxdxxax22cos1sincoscossinsin22222222令,CxaxaxaCtata222222arcsin22sin421.(35)12xxdx;解 CxCtdttdtttttxxxdx1arccostansectansec1sec12令.或 Cxxdxdxxxxxdx1arccos111

17、111112222.(36)32)1(xdx;解 Cttdttdttxxdxsincostan)1(tan1tan)1(3232令Cxx12.(37)dxxx92;解 tdttdtttxdxxx222tan3)sec3(sec39sec9sec39令 CxxCttdtt3arccos393tan3)1cos1(322.(38)xdx21;解 CxxCttdtttdtttxxdx)21ln(2)1ln()111(11221令.(39)211xdx;解 dttdtttdtttxxdx)2sec211()cos111(coscos11sin1122令 CxxxCtttCtt211arcsincos1

18、sin2tan.(40)21 xxdx.解 dttttttttdttttxxxdxcossinsincossincos21coscossin1sin12令 Ctttttdttdt|cossin|ln2121)cos(sincossin12121 Cxxx|1|ln21arcsin212.习题 43 求下列不定积分:1.xdxxsin;解 Cxxxxdxxxxxdxdxxsincoscoscoscossin.2.xdxln;解 Cxxxdxxxxxdxxxdxlnlnlnlnln.3.xdxarcsin;解 xxdxxxdxarcsinarcsinarcsin dxxxxx21arcsin Cx

19、xx21a r c s i n.4.dxxex;解 dxexexdedxxexxxx CxeCexexxx)1(.5.xdxx ln2;解 xdxxxxdxxdxxln31ln31ln31ln3332 Cxxxdxxxx332391ln3131ln31.6.xdxexcos;解 因为 xdxexexdexexdexdxexxxxxxsinsinsinsinsincos xxxxxxdexexexdexecoscossincossin xdxexexexxxcoscossin,所以 CxxeCxexexdxexxxx)cos(sin21)cossin(21cos.7.dxxex2sin2;解 因

20、为 xxxxdexxexdedxxe22222cos22cos22cos22sin 2sin82cos22cos42cos22222xdexedxxexexxxx xxxdexxexe2222sin82sin82cos2 dxxexexexxx2sin162sin82cos2222,所以 Cxxedxxexx)2sin42(cos1722sin22.8.dxxx2cos;解 Cxxxdxxxxxxddxxx2cos42sin22sin22sin22sin22cos.9.xdxx arctan2;解 dxxxxxxdxxdxx233321131arctan31arctan31arctan 223

21、2223)111(61arctan31161arctan31dxxxxdxxxxx Cxxxx)1ln(6161arctan31223.10.xdxx2tan 解 xxdxxdxxdxxdxxxxdxxtan21sec)1(sectan2222 Cxxxxxdxxxx|cos|lntan21tantan2122.11.xdxx cos2;解 xxdxxxdxxxxxdxxdxxcos2sin2sinsinsincos2222 Cxxxxxxdxxxxxsin2cos2sincos2cos2sin22.12.dttet 2;解 dtetetdedttetttt2222212121 CteCete

22、ttt)21(214121222.13.xdx2ln;解 xdxxxdxxxxxxxdxln2ln1ln2lnln222 Cxxxxxdxxxxxxx2ln2ln12ln2ln22.14.xdxxxcossin;解 xdxxxxxdxdxxxdxxx2cos412cos412cos412sin21cossin Cxxx2sin812cos41.15.dxxx2cos22;解 xdxxxxxxdxxdxxxdxxxsinsin2161sin2161)cos1(212cos2323222 xdxxxxxxxxdxxxcoscossin2161cossin21612323 Cxxxxxxsincos

23、sin216123.16.dxxx)1ln(;解 dxxxxxdxxdxxx1121)1ln(21)1ln(21)1ln(222 dxxxxx)111(21)1ln(212 Cxxxxx)1ln(212141)1ln(2122.17.xdxx2sin)1(2;解 xdxxxxxdxxdxx22cos212cos)1(212cos)1(212sin)1(222 xxdxx2sin212cos)1(212 xdxxxxx2sin212sin212cos)1(212 Cxxxxx2cos412sin212cos)1(212.18.dxxx23ln;解 xdxxxxxdxxxxxddxxx223333

24、23ln13ln1ln1ln11lnln xdxxxxxxxdxx22323ln13ln3ln11ln3ln1 xxdxxxxdxxxxxxx1ln6ln3ln1ln16ln3ln123223 dxxxxxxxx22316ln6ln3ln1 Cxxxxxxx6ln6ln3ln123.19.dxex3;解 ttxdetdtettxdxe223333令 tttttdeetdtteet636322 dteteetttt6632 Ceteetttt6632 Cxxex)22(33323.20.xdxlncos;解 因为 dxxxxxxxdx1lnsinlncoslncos dxxxxxxxxxdxxx

25、1lncoslnsinlncoslnsinlncos xdxxxxxlncoslnsinlncos,所以 Cxxxxdx)lnsinln(cos2lncos.21.dxx2)(arcsin;解 dxxxxxxdxx22211arcsin2)(arcsin)(arcsin 221arcsin2)(arcsinxxdxx dxxxxx2arcsin12)(arcsin22 Cxxxxx2arcsin12)(arcsin22.22.xdxex2sin.解 xdxeedxxexdxexxxx2cos2121)2cos1(21sin2,而 dxxexexdexdxexxxx2sin22cos2cos2c

26、os xdxexexedexxexxxxx2cos42sin22cos2sin22cos,Cxxexdxexx)2sin22(cos512cos,所以 Cxxeexdxexxx)2sin22(cos10121sin2.习题 44 求下列不定积分:1.dxxx33;解 dxxxxxdxxxdxxx327)93)(3(327273233 dxxdxxx3127)93(2 Cxxxx|3|ln279233123.2.dxxxx103322;解 Cxxxxdxxdxxxx|103|ln)103(1031103322222.3.dxxxxx3458;解 dxxxxxdxxxdxxxxx3223458)1

27、(8 dxxdxxdxxxxx13148213123 Cxxxxxx|1|ln3|1|ln4|ln8213123.4.dxx133;解 dxxxxxxxdxxxxxdxx)11231122111()1211(132223 )21()23()21(123)1(1121|1|ln2222xdxxxdxxx Cxxxx312arctan31|1|ln2.5.)3)(2)(1(xxxxdx;解 dxxxxxxxxdx)331124(21)3)(2)(1(Cxxx|)1|ln|3|ln3|2|(ln21.6.dxxxx)1()1(122;解 dxxxxdxxxx)1(111211121)1()1(122

28、2 Cxxx11|1|ln21|1|ln21 Cxx11|1|ln212.7.dxxx)1(12;解 Cxxdxxxxdxxx)1ln(21|ln)11()1(1222.8.)(1(22xxxdx;解 dxxxxxxxxdx)112111211()(1(222 dxxxxx1121|1|ln21|ln2 dxxdxxxxx11211241|1|ln21|ln22 Cxxxxarctan21)1ln(41|1|ln21|ln2.9.)1)(1(22xxxdx;解 dxxxxxxxxxdx)111()1)(1(2222 )1ln(21112111221222xdxxxxxx dxxxxxx1121

29、)1ln(21|1|ln21222 Cxxxx312arctan33)1ln(21|1|ln2122.10.dxx114;解 dxxxxxdxx)12)(12(111224 dxxxxdxxxx12214212214222 dxxxxdxxxx1222)22(21421222)22(214222 )1212(4112)12(12)12(82222222xxdxxxdxxxxxdxxxxd Cxxxxxx)12arctan(42)12arctan(42|1212|ln8222.11.dxxxx222)1(2;解 dxxxdxxxxdxxxx11)1(1)1(2222222 dxxxdxxxdxx

30、xx11)1(123)1(122122222 dxxxdxxxxx11)1(12311212222,因为 )312arctan(32)312()312(11321122xxdxdxxx,而 dxxdxxx22222)23()21(1)1(1 由递推公式 )()32()()1(21)(122122222nnnaxdxnaxxnaaxdx,得 dxxdxxx22222)23()21(1)1(1 312arctan323211231)1121()23(212222xxxxxxdxxxx,所以 dxxxx222)1(2Cxxxxxxx312arctan32312arctan3211221112122

31、Cxxxx312arctan34112.12.xdx2sin3;解 xdxdxxxdxtan3tan41cos41sin3222 Cxxdx3tan2arctan321tan)23(tan14122.13.dxxcos31;解)2sec1(2cos)2(2cos121cos31222xxxdxdxdxx Cxxxd22tanarctan212tan22tan2.或 duuuuxudxx221212312tancos31令 CxCuduu22tanarctan212arctan21)2(122.14.dxxsin21;解)2cot2(csc2sin)2(2cos2sin22sin2122xxxx

32、dxxdxdxx 222)23()212(cot)212(cot12cot2cot)2(cotxxdxxxd Cx312cot2arctan32.或 duuuuxudxx221212212tansin21令 duuduuu222)23()21(111 CxCu312tan2arctan32312arctan32.15.xxdxcossin1;解 Cxxxdxxdxxxdx|2tan|ln2tan1)2(tan)2tan1(2cos21cossin12.或 duuuuuuxuxxdx2222121112112tancossin1令 CxCuduu|12tan|ln|1|ln11.16.5coss

33、in2xxdx;解 duuuduuuuuuxuxxdx2231125111412tan5cossin222222令 CxCuduu512tan3arctan51513arctan51)35()31(13122.或 duuuuuuxuxxdx2222125111412tan5cossin2令 duuduuu222)35()31(1312231 CxCu512tan3arctan51513arctan51.17.dxx3111;解 duuuduuuuxdxx)111(33111111233令 CxxxCuuu)11ln(313)1(23|1|ln332333322.18.dxxx11)(3;解 C

34、xxxdxxxdxxx232233221 1)(11)(.19.dxxx1111;解 duuuuduuuuxdxxx)122(221111111令 Cuuu|)1|ln2221(22 Cxxx)11ln(414)1(.20.4xxdx;解 duuuuuxxxdx324441令 Cuuuduuu|1|ln442)111(42 Cxxx)1ln(4244.21.xdxxx11;解 令uxx11,则2211uux,duuudx22)1(4,duuuduuuuuuxdxxx)1111(2)1(41111222222 Cuuuarctan2|11|ln Cxxxxxx11arctan2|1111|ln.

35、22.342)1()1(xxdx.解 令uxx311,则1133uux,232)1(6uudx,代入得 CxxCuduxxdx334211232323)1()1(.总习题四 求下列不定积分(其中 a,b 为常数):1.xxeedx;解 Ceedeedxeeeedxxxxxxxxx|11|ln2111122.2.dxxx3)1(;解 Cxxdxxdxxdxxx2323)1(12111)1(1)1(1)1(.3.dxxax662(a0);解 Caxaxaxdxadxxax|ln61)()()(1313333332323662.4.dxxxxsincos1;解 Cxxxxdxxdxxxx|sin|l

36、n)sin(sin1sincos1.5.dxxxlnln;解 Cxxxdxxxxxxxxddxxxlnlnlnln1ln1lnlnlnlnlnlnlnlnln.6.dxxxx4sin1cossin;解 Cxxdxxdxxdxxxx222244sinarctan21)(sin)(sin1121sinsin1sinsin1cossin.7.xdx4tan;解 xxdxxdxxxdxtansintantancossintan22244 xdxxxdxxtan)1tan11(tantan1tantan2224 cxxxcxxxtantan31tanarctantantan3133.8.xdxxx3si

37、n2sinsin;解 xdxxxxdxxx3sin)cos3(cos213sin2sinsin xdxxxdxx3sincos213sin3cos21 dxxxxxd)2sin4(sin41)3(cos3cos61 Cxxx2cos814cos1613cos1212.9.)4(6xxdx;解 Cxxdxxxxxxdx)4ln(241|ln41)41(41)4(6656.10.)0(adxxaxa;解 dxxaxdxxaaduxaxadxxaxa2222221 Cxaaxa22arcsin.11.)1(xxdx;解 CxxCxxxdxxxdx)1ln(2)(1ln(2)(112)1(22.12.

38、xdxx2cos;解 xxdxdxxxxxdxx2sin4141)2cos(21cos22 Cxxxxxdxxxx2cos812sin41412sin412sin414122.13.bxdxeaxcos;解 因为 dxbxeabbxeabxdeabxdxeaxaxaxaxsincos1cos1cos dxbxeabbxeabbxeadebxabbxeaaxaxaxaxaxcossincos1sincos12222,所以 Cbxeabbxeabaabxdxeaxaxax)sincos1(cos2222 Cbxbbxaebaax)sincos(122.14.xedx1;解 duuuduuuduue

39、edxxx)1111(112)1ln(11122令.ceecuuxx1111ln|11|ln.15.122xxdx;解 Cttdttdtttttxxxdxsincostansectansec1sec1222令 Cxx12.16.2/522)(xadx;解 tdtatataxxadxcos)cos(1sin)(52/522令 tdtadttatan)1(tan1cos112444 Ctatatan1tan31434 Cxaxaxaxa224322341)(31.17.241 xxdx;解 tdttttxxxdx2424secsectan1tan1令 tdttdtttsinsincossincos

40、4243 Ctttdttsin1sin31sin)sin1sin1(324 Cxxxx233213)1(.18.dxxxsin;解 tdtttdttttxdxxxsin22sinsin2令 tdtttttdt2cos2cos2cos222 tdtttttttdttsin4sin4cos2sin4cos222 Ctttttcos4sin4cos22 Cxxxxxcos4sin4cos2.19.dxx)1ln(2;解 dxxxxxxdxx22212)1ln()1ln(dxxxx)111(2)1ln(22 Cxxxxarctan22)1ln(2.20.dxxx32cossin;解 xdxxxxdxx

41、dxxxtan)1tantan(tantancossincossin2232 Cxx)1ln(tan21tan2122.21.dxxarctan;解 xdxxxxdxx11arctanarctan xdxxx)111(arctan Cxxxxarctanarctan Cxxxarctan)1(.22.dxxxsincos1;解 Cxxxdxdxxxxdxxx|2cot2csc|ln222csc22cos2sin22cos2sincos1.23.dxxx283)1(;解 Cxxxdxxdxxxarctan12141)1(141)1(484428283.提示:已知递推公式 )()32()()1(2

42、1)(122122222nnnaxdxnaxxnaaxdx.24.dxxxx234811;解 dtttttxdxxxxdxxxx234123412322444884811令 dtttdtttt)11241(41)23231(412 Cttt|1|ln41|2|ln41 Cxxx21ln414444.25.416 xdx;解 dxxxdxxxxdx)4141(81)4)(4(11622224 Cxxx)2arctan21|22|ln41(81 Cxxx2arctan161|22|ln321.26.dxxxsin1sin;解 dxxxxdxxxxdxxx222cossinsinsin1)sin1(

43、sinsin1sin Cxxxdxxxxtansec)cos11cossin(22.27.dxxxxcos1sin;解 dxxxdxxxdxxxxdxxxx2cossin212cos212cos2sincos1sin222 dxxxxd2tan2tan Cxxdxxdxxxx2tan2tan2tan2tan.28.dxxxxxex23sincossincos;解 xdxxexdxexdxxxxxexxxsectancoscossincossinsin23sin xdexdxexxsecsinsinsin xxxxdeexxdesinsinsinsecsec xdxexexdxexexxxxco

44、ssecsecsinsinsinsin Cexxexxsinsinsec.29.dxxxxx)(33;解 dtttdtttttttxdxxxxx)111(66)()(52362633令 CxxCtt66)1(ln1ln6.30.2)1(xedx;解 dttttdtttteedxxx)1111(1111)1(222令 Cttt1ln)1ln(Ceexxx11)1ln(.31.dxeeeexxxx1243;解)()(1111222243xxxxxxxxxxxxeedeedxeeeedxeeee Ceexx)arctan(Cx)sh2arctan(.32.dxexexx2)1(;解 11)1()1(

45、)1(22xxxxxexdedexdxexe xxxxxxdeeeexdxeex)1(11111 xxxxdeeeex)111(1 Ceeexxxx)1ln(ln1 Ceexexxx)1ln(1.33.dxxx)1(ln22;解 dxxxxxxxdxxx)1(ln)1(ln)1(ln222222 dxxxxxxxx22221)1ln(2)1(ln 22221)1ln(2)1(lnxdxxxxx dxxxxxxxxxx)1ln(12)1ln(12)1(ln222222 dxxxxxxx2)1ln(12)1(ln2222 Cxxxxxxx2)1ln(12)1(ln2222.34.dxxx2/32)

46、1(ln;解 因为 CxxCttdttdtttxdxx2232/321sincossecsec1tan)1(1令所以 dxxxxxxxxxxddxxx111ln)1(ln)1(ln2222/32 Cxxxxx)1ln(1ln22.35.xdxx arcsin12;解 dtttttdtttxxdxx)2cos(21cossinarcsin122令 tdttttttt2sin412sin41412sin414122 Ctttt2cos812sin41412 122241arcsin121)(arcsin41Cxxxxx.36.dxxxx231arccos;解 2222231arccos1arcco

47、s1arccosxxdxdxxxxxdxxxx dxxxxxxx)arccos(1arccos12222 dxxxxxxxxx)11arccos2(1arccos122222 dxxxdxxxxxx2222arccos12arccos1 32322)1(arccos3231arccos1xxdxxxx dxxxxxxxx)1(32arccos)1(3231arccos1232322 Cxxxxxxxx3323229232arccos)1(3231arccos1 Cxxxxx)6(91arccos)1(131222.37.dxxxsin1cot;解 xdxxxdxxdxxxsin)sin11si

48、n1(sin)sin1(sin1sin1cot ln|sin x|ln|1sin x|Cln|csc x1|C.38.xxdxcossin3;解 xdxxxdxxxxdxxxxdxcotcos1cotcotcossincoscotcossin1cossin223 122sin21|tan|lncot21|cot|lncot)cotcot1(CxxCxxxdxx.39.xxdxsin)cos2(;解 令2tanxu,则 duuuuduuuuuuxxdx)3(11212)112(1sin)cos2(222222 Cuuduuduuu|ln31)3ln(31131323122 Cxx|2tan32tan|ln313.40.dxxxxxcossincossin;解 dxxxxdxxxxduxxxxxxdxxxxx1cos2sincos1sin2cossincossincossin)cos(sincossincossin222222 xdxxxdxxcos1cos2cossin1sin2sin2222 xdxxdxcos)1cos211(21sin)1sin211(2122 Cxxxxxx|1cos21cos2|ln21cos21|1sin21sin2|ln21sin21 Cxxxx|1cos21sin2|ln2)cos(sin21.