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第四章不定积分习题A(作业题)1.(1)原式=cxdxx3231)23(43)23(21(2)原式=cxxdxdxxxcos2cos)(cos)(cossin2323(3)原式=xdxxxxdxarcsin2arcsin2arcsin2cxxxdxxxxxx12arcsin22112arcsin22.cxdxxyln13)(2ey,得1c,故1lnxy习题B(练习题)1.填空题(1)cy(2)cxcos2(3)cex(4)xdxxxcossin)ln(sin(5)cx221(6)2cos2xx(7))2(xf(8)cx232)1(31(9)xarcsin(10)cxx2sincos(11)67213123xx,原题改为dxxxdxxf)()('2(12)cxx2lnln2(13)cxx2)1ln(2.选择题(1)A(2)C(3)B(4)B(5)A(6)B(7)B(8)B(9)、D(10)、B(11)、C3、计算下列不定积分:(1)cxxdxxxxdxxxxsectan)tansec(sec)tan(secsec2(2)cxxxdxdxx1110210tan111tantansectan(3)cxxdxxxdxxdxxxdxxxdxxxx)ln(lnlnlnlnlnln1lnlnlnln1lnlnlnln1lnlnln1lnlnln1(4)cxxxxdxxdxxxxxdxxxxxdxxxsincosln)cos(sinsincos1sincossincoscossin1cossin1tan1tan1(5)cttdtdtttcos2sin2sin(6)cexedexdxeedxeeedxexxxxxxxxx|1|ln)1(11)11(1111(7)cxxdxdxxx2)2tan(ln212tanln2tanlnsin2tanln(8)cxxxxdxxxxxdxxxxxdxxdxxdxxx2tan212tan)1(sec212tan)tantan(21tan21tantansectan2222222(9)cxxdxdxxx)21(2arcsin)2()21(411)1(12(10)cxxxxddxxxxln1ln1)ln(ln12(11)dxxx1arcsin设txarcsin,则tx2sin.得到原式为:cxxxcttttdtttttdtdtttdtttt2arcsin12sin2cos2)coscos(2cos2sin2cossin2cos(12)cxxxxdxxdxxxx|103|ln)103(1031103322222(13)ceeceedeeedeedxeexxxxxxxxxxx|11|ln21|1|ln|1|ln21)1111(211112)((14)cxxxxdxxdxxxx252323)ln(52)ln()ln()ln1()ln((15)txdxx2211设原式cxxcttdttdttt|12|ln2|1|ln)111(1(16)311x3132txtdxxx设cxxcttdttt353522)13(272)13(452272452)1(92原式(17)xttxxxdx1arccossec12设原式=cxxddttdtt1arccos1arccossectansec1(18)dxxx21设txtan则xtarctancxxxxcttdtttttdttdttdxxx|11|ln1|2tan|lnsecsin1sintantansin1tantansec1222原式(19)cxxxdxxdxxx21arcsin)1(421)1(423222(20)cexdedxxxexxx22211211321111)1((21)cxxxdxxxxxdxxxxxdxdxx3323333291ln3131ln31ln3131ln31lnln(22)cxxxxxdxxxdxxxxxxxxdxxdxxarctanarctan)111(arctan211arctanarctanarctanarctan(23)cxecexexdxexexexdxxexxdexdxxexxxxxxxxxx221)1(21)1()1(2121)1(21)1()2()1(2(24)cxxxxdxxxdxxx122ln51))12ln(21)2ln(21(52)121)2(21(5223212(25)cxxxcxxxdxxxdxxdxxdxxxxxDCBADBADCBDCBACxDCxxBxAxxdxxx11ln21221|1|ln41|1|ln2111211211121)1(121)121121)1(21(0,21,21002020B11)1()1()1(1)1()1(1222222222222原式于是所以得令(26)cxxxxxdxxxddxxxxdxxxxxxdxxxxx|32|ln532(3215)32225(3222)32(5321312522222222)(27)cxxxxdxxdxdxxdxxxdxdxxx23arctan4)136(ln)3(4)3(18]4)3[(4)3(1214)3(84)3(343)-(x5x1365x22222222(28)cxxcxxdxxxxCBAAcBAxxxxCBxxAdxxxx|)1(|ln|1|ln72||ln)121(0,2,1101)1(11)1(1727776777677原式则令(29)cxxxxxddxxxdxxxdxdxxxxdxxxxdxx2sin4814sin641162sin2sin161)4cos1(1612cos2sin812sin8122cos142sincoscossincossin3222222242(30)cxxxdxxdxxxdxxdxxxcos1coscos)cos11(coscos1coscoscossincossin22222234、求过点)0,21(且满足关系式11arcsin'2xyxy的曲线方程。解:1)'arcsin(1arcsin'2xyxyxy于是对等式二端求积分得:cxxydxdxxyarcsin1)'arcsin(且方程过2166210),0,21(c带入方程求得216arcsinxxy所以曲线方程为5、已知)(xf的一个原函数为xxxsin1sin,求dxxfxf)(')(解:22)sin1(sincos)'sin1sin()('xxxxxxxxfCxxxCxfxdfxfdxxfxf222)sin1(sin21)(21)()()(')(6、已知曲线)(xfy在任意点处的切线斜率为632xax,且1x时211y为极大值,试确定)(xf,并求)(xf的极小值。解:Cxxxadxxaxxf6233)63()(232由于)(xf在1x时有极大值故0)1('f且211)1(f于是063)1('af解得3a由1x时211y可知211)1(f求得2C于是2623)(23xxxxf令0)2)(1(3633)('2xxxxxf求得1x或2x,当x在2的左侧邻近时,;0)('xf当x在2的右侧邻近时,;0)('xf于是,当2x时函数取到极小值,且极小值8)2(f7、已知曲线)(xfy过点)21,0(且其上任一点),(yx处的切线斜率为)1ln(2xx,求曲线方程。解:cxxxcxxxxxdxdxxxdxxxxxxdxxxdxxdxxxxf222122222222222222222222221)1ln()1(21)]1ln()1ln([21)]1(111)1ln([21]1)1ln([21])1ln()1ln([21)1ln(21)1ln()(由题知:21)0(f故21c于是得曲线方程:2121)1ln()1(21)(222xxxxf习题C(提高题)1、计算xdxex4sin2sin52解:cedexdxexxx2sin52sin52sin52224sin2、计算dxxx)1(110解:令yx1,则cxcyydydyyydxxx)11ln(101)1ln(101)1(111011)1(110101010109103、计算dxexxexxxx)13()(22解:Cexxexxdexxdxexxexxxxxxx2322222])[(32)()(2)13()(4、计算dxxx)1,,max(23解:由图像可知]1,1[1),1(4341)1,(3231)1,,max(43,32,1411141411131113131,1]1,1[1),1(41)1,(31)1,,max(4332232133233224331331132324313223xCxdxxCxdxxxCxdxxdxxxCCCCCCCCCCxdxCCxdxxxCCCCxdxCCxdxxxxCxdxxCxdxxxCxdxxdxxx则令时,当时当5、设)(xyy是由xyyx333所确定的隐函数,试求dxxyxyy2)1(解:Cxydxxydxxyxxxyydxxyxyydxxyxyyxyxyyxyyx2222222223321)'()1()1('3二边同时求导得对(6)设010sin)(xexxxfx,求dxxf)1(解:111)1sin()1()1(xexxxfx于是:1)1(11)1cos()1sin()1(,11C)1(1)1sin(11)1(1)1cos()1sin()1()1()1(21122)1(12)1()1(1xCxedxexCxdxxdxxfCCCCCCCdxeCdxxxxCxedxexCxdxxdxxfxxxxx设时,当
本文标题:[09.11]工科高数第四章答案
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