高等数学第六版课后习题答案.pdf

第四章习题 4−11. 求下列不定积分:(1)∫dx x21;解CxCxdxxdxx+−=++−==+−−∫∫1 1211122 2.(2)∫dxxx;解CxxCxdxxdxxx+=+ +==+∫∫2123 23521231.(3)∫dx x1;解CxCxdxxdx x+=+ +−==+−−∫∫2 12111121 21 .(4)∫dxxx32;解CxxCxdxxdxxx+=+ +==+∫∫33137 37 32 1031371.(5)∫dx xx21;解C xxCxdxxdx xx+⋅−=+ +−==+−−∫∫1 2312511125 252.(6)dxxmn∫;解CxmnmCxmndxxdxxmnm mn mn mn++=+ +==++∫∫111.(7)∫dxx35;解Cxdxxdxx+==∫∫433 4555.(8)∫+−dxxx)23(2;解Cxxxdxdxxdxxdxxx++−=+−=+−∫∫∫∫223 3123)23(2322.(9)∫ ghdh2(g是常数);解CghCh gdhh gghdh+=+⋅==∫∫−22 2121221 21 .(10)∫−dxx2)2(;解Cxxxdxdxxdxxdxxxdxx++−=+−=+−=−∫∫∫∫∫423144)44()2(23222.(11)∫+dxx22) 1(;解Cxxxdxdxxdxxdxxxdxx+++=++=++=+∫∫∫∫∫35242422 32 512) 12() 1(.(12)dxxx∫−+) 1)(1(3;解∫∫∫∫∫∫−+−=−+−=−+dxdxxdxxdxxdxxxxdxxx23 21 2323) 1() 1)(1(Cxxxx+−+−=25 23 3 52 32 31.(13)∫−dx xx2)1 (;解Cxxxdxxxxdx xxxdx xx++−=+−=+−=−∫∫∫−25 23 21 23 21 212252 342)2(21)1 (.(14)∫+++dxxxx 1133224 ;解Cxxdx xxdxxxx++=++=+++∫∫arctan)113(11333 22 224 .(15)∫+dxxx221;解∫∫∫+−=+−=+−+=+Cxxdxxdxxxdxxxarctan)111 (111122222 .(16)∫+dxxex)32(;解Cxedxxdxedxxexxx++=+=+∫∫∫||ln32132)32(.(17)∫ −−+dx xx) 1213( 22;解∫∫∫+−= −−+= −−+Cxxdx xdxxdx xxarcsin2arctan3 112113) 1213( 2222.(18)dx xeex x∫− −)1 (;解Cxedxxedx xeexxx x+−=−=−∫∫−− 21 21 2)()1 (.(19)∫dxexx3;解CeCeedxedxexxx xxx++=+==∫∫13ln3 )3ln()3()3(3.(20)∫⋅−⋅dxxxx32532;解CxCxdxdxxxx xxx +−−=+−=−=⋅−⋅∫∫)32(3ln2ln5232ln)32( 52])32(52[32532.(21)∫−dxxxx)tan(secsec;解∫∫+−=−=−Cxxdxxxxdxxxxsectan)tansec(sec)tan(secsec2.(22)∫dxx 2cos2;解Cxxdxxdxxdxx++=+=+=∫∫∫)sin(21)cos1 (21 2cos1 2cos2.(23)∫+dxx2cos11;解∫∫+==+Cxdxxdxxtan21cos21 2cos112.(24)∫−dxxxx sincos2cos;解∫∫∫+−=+=−−=−Cxxdxxxdxxxxxdxxxxcossin)sin(cossincossincos sincos2cos22 .(25)∫dxxxx22sincos2cos;解∫∫∫+−−=−=−=Cxxdxxxdxxxxxdxxxxtancot)cos1 sin1(sincossincos sincos2cos22222222.(26)∫−dxxx x)11 (2;解∫⎟⎠⎞⎜⎝⎛−dxxxx211∫++=−=−−Cxxdxxx41 47 45 43 474)(.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y=f(x), 则由题意得xxfy1)(=′=′,所以Cxdxxy+==∫||ln1.又因为曲线通过点(e2, 3), 所以有=3−2=13=f(e2)=ln|e2|+C=2+C, C=3−2=1. 于是所求曲线的方程为y=ln|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s=s(t), 则s′=v=3t2,Ctdtts+==∫323.因为当t=0 时,s=0, 所以C=0. 因此位移函数为s=t3.(1)在 3 秒后物体离开出发点的距离是s=s(3)=33=27.(2)由t3=360, 得物体走完 360m 所需的时间11. 73603≈=ts.4. 证明函数xe221,exshx和exchx都是xxex shch −的原函数.证明x xxxxxxxxeee eeeee xxe222shch==−−+=−−−−.因为xxee22)21(=′, 所以xe221是xxex shch −的原函数.因为 (exshx)′=exshx+exchx=ex(shx+chx)xxxxxxeeeeee2)22(=++−=−−,所以exshx是xxex shch −的原函数.因为 (exchx)′=exchx+exshx=ex(chx+shx)xxxxxxeeeeee2)22(=−++=−−,所以exchx是xxex shch −的原函数.习题 4—21. 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(41+=xddx:(1)dx= d(ax);解dx= a1d(ax).(2)dx=d(7x−3);解dx=71d(7x−3).(3)xdx=d(x2);解xdx=21d(x2).(4)xdx=d(5x2);解xdx=101d(5x2).(5))1 ( 2xdxdx−=;解)1 ( 212xdxdx−−=.(6)x3dx=d(3x4−2);解x3dx=121d(3x4−2).(7)e2xdx=d(e2x);解e2xdx=21d(e2x).(8))1 ( 22xxeddxe−−+=;解)1 ( 2 22xxeddxe−−+−=.(9))23(cos 23sinxdxdx=;解)23(cos 3223sinxdxdx−=.(10)|)|ln5( xdxdx=;解|)|ln5( 51xdxdx=.(11)|)|ln53( xdxdx−=;解|)|ln53( 51xdxdx−−=.(12))3(arctan 912xdxdx=+;解)3(arctan 31912xdxdx=+.(13))arctan1 ( 12xd xdx−= −;解)arctan1 ( ) 1( 12xd xdx−−= −.(14))1( 122xd xxdx−= −.解)1( ) 1( 122xd xxdx−−= −.2. 求下列不定积分(其中a,b,ω,ϕ均为常数):(1)∫dtet5;解Cexdedtexxt+==∫∫555 51551.(2)∫−dxx3)23(;解Cxxdxdxx+−−=−−−=−∫∫433)23(81)23()23(21)23(.(3)∫−dxx211;解Cxxdxdxx+−−=−−−=−∫∫|21|ln21)21 (211 21 211.(4)∫ −332xdx;解CxCxxdx xdx+−−=+−⋅−=−−−= −∫∫−32 32 313)32(21)32(23 31)32()32(3132.(5)∫−dxeaxbx )(sin;解Cbeaxabxdebaxdaxadxeaxbx bx bx +−−=−=−∫∫∫cos1)()(sin1)(sin.(6)∫dt ttsin;解∫∫+−==Cttdtdt ttcos2sin2sin.(7)∫⋅xdxx210sectan;解∫⋅xdxx210sectanCxxxd+==∫1110tan111tantan.(8)∫xxxdx lnlnln;解Cxxdxxdxxxxxdx+===∫∫∫|lnln|lnlnlnlnln1lnlnlnln1 lnlnln.(9)∫ +⋅+dx xxx 2211tan;解∫+⋅+dx xxx 2211tan222 221 1cos1sin11tanxd xxxdx+ ++=++=∫∫Cxxd x++−=+ +−=∫|1cos|ln1cos 1cos1222.(10)∫xxdx cossin;解Cxxdxdxxx xxdx+===∫∫∫|tan|lntantan1 tansec cossin2 .(11)∫−+dxeexx1;解∫−+dxeexx1Cedeedxeexx xxx +=+=+=∫∫arctan11122.(12)∫−dxxex2;解.21)(212222Cexdedxxexxx+−=−−=−−−∫∫(13)∫⋅dxxx)cos(2;解Cxxdxdxxx+==⋅∫∫)sin(21)()cos(21)cos(2222.(14)∫ −dx xx232;解CxCxxdxdx xx+−−=+−−=−−−= −∫∫−221 2221 223231)32(31)32()32(6132.(15)∫−dxxx4313;解∫∫+−−=−−−=−Cxxdxdxxx|1|ln43)1 (11 431344 443 .(16)∫++dttt))sin((cos2ϕωϕω;解Cttdtdttt++−=++−=++∫∫)(cos31)cos()(cos1)sin()(cos322ϕωωϕωϕωωϕωϕω.(17)∫dxxx3cossin;解CxCxxxddx xx+=+=−=−−∫∫223 3sec21cos21coscoscossin.(18)∫ −+dx xxxx3cossincossin;解)sincos( cossin1cossincossin33xxd xxdx xxxx+− −= −+∫∫Cxxxxdxx+−=−−=∫−32 31 )cos(sin23)cos(sin)cos(sin.(19)∫ −−dx xx2491;解dx xxdx xdx xx∫∫∫−− −= −−22249491491)49( 491 81)32()32(11 21222xd xxdx− −+−=∫∫Cxx+−+=24941 32arcsin21.(20)∫+dxxx239;解Cxxxdxxdxxdxxx++−=+−=+=+∫∫∫)]9ln(9[21)()991 (21)(9219222 22 2223 .(21)∫−dxx1212;解∫∫∫+− −= +−=−dx xxdx xxdxx) 121121(21) 12)(12(11212∫∫+ +−− −=) 12( 121221) 12( 121221xd xxd xC xxCxx+ +−=++−−=| 1212|ln 221|12|ln 221|12|ln 221.(22)∫−+dxxx)2)(1(1;解CxxCxxdxxxdxxx++−=++−−=+−−=−+∫∫|12|ln31|1|ln|2|(ln31)11 21(31 )2)(1(1.(23)∫xdx3cos;解Cxxxdxxdxxdx+−=−==∫∫∫3223sin31sinsin)sin1 (sincoscos.(24)∫+dtt)(cos2ϕω;解Cttdttdtt+++=++=+∫∫)(2sin41 21)](2cos1 [21)。