搜索
习题4−11.求下列不定积分:(1)∫dxx21;解CxCxdxxdxx+−=++−==+−−∫∫112111222.(2)∫dxxx;解CxxCxdxxdxxx+=++==+∫∫212323521231.(3)∫dxx1;解CxCxdxxdxx+=++−==+−−∫∫21211112121.(4)∫dxxx32;解CxxCxdxxdxxx+=++==+∫∫3313737321031371.(5)∫dxxx21;解CxxCxdxxdxxx+⋅−=++−==+−−∫∫12312511125252.(6)dxxmn∫;解CxmnmCxmndxxdxxmnmmnmnmn++=++==++∫∫111.(7)∫;dxx35解Cxdxxdxx+==∫∫4334555.(8)∫;+−dxxx)23(2解Cxxxdxdxxdxxdxxx++−=+−=+−∫∫∫∫2233123)23(2322.(9)∫ghdh2(g是常数);解CghChgdhhgghdh+=+⋅==∫∫−22212122121.(10)∫;−dxx2)2(解Cxxxdxdxxdxxdxxxdxx++−=+−=+−=−∫∫∫∫∫423144)44()2(23222.(11)∫;+dxx22)1(解Cxxxdxdxxdxxdxxxdxx+++=++=++=+∫∫∫∫∫3524242232512)12()1(.(12)dxxx∫−+)1)(1(3;解∫∫∫∫∫∫−+−=−+−=−+dxdxxdxxdxxdxxxxdxxx23212323)1()1)(1(Cxxxx+−+−=25233523231.(13)∫−dxxx2)1(;解Cxxxdxxxxdxxxxdxxx++−=+−=+−=−∫∫∫−2523212321212252342)2(21)1(.(14)∫+++dxxxx1133224;解Cxxdxxxdxxxx++=++=+++∫∫arctan)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Cxedxxedxxeexxxx+−=−=−∫∫−−21212)()1(.(19)∫;dxexx3解CeCeedxedxexxxxxx++=+==∫∫13ln3)3ln()3()3(3.(20)∫⋅−⋅dxxxx32532;解CxCxdxdxxxxxxx+−−=+−=−=⋅−⋅∫∫)32(3ln2ln5232ln)32(52])32(52[32532.(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(cossincossincossincos2cos22.(25)∫dxxxx22sincos2cos;解∫∫∫+−−=−=−=Cxxdxxxdxxxxxdxxxxtancot)cos1sin1(sincossincossincos2cos22222222.(26)∫−dxxxx)11(2;解∫⎟⎠⎞⎜⎝⎛−dxxxx211∫++=−=−−Cxxdxxx41474543474)(.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xxexshch−的原函数.证明xxxxxxxxxeeeeeeeexxe222shch==−−+=−−−−.因为xxee22)21(=′,所以xe221是xxexshch−的原函数.因为(exshx)′=exshx+exchx=ex(shx+chx)xxxxxxeeeeee2)22(=++−=−−,所以exshx是xxexshch−的原函数.因为(exchx)′=exchx+exshx=ex(chx+shx)xxxxxxeeeeee2)22(=−++=−−,所以exchx是xxexshch−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(222xxeddxe−−+−=.(9))23(cos23sinxdxdx=;解)23(cos3223sinxdxdx−=.(10)|)|ln5(xdxdx=;解|)|ln5(51xdxdx=.(11)|)|ln53(xdxdx−=;解|)|ln53(51xdxdx−−=.(12))3(arctan912xdxdx=+;解)3(arctan31912xdxdx=+.(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,ω,ϕ均为常数):(1)∫;dtet5解Cexdedtexxt+==∫∫55551551.(2)∫;−dxx3)23(解Cxxdxdxx+−−=−−−=−∫∫433)23(81)23()23(21)23(.(3)∫−dxx211;解Cxxdxdxx+−−=−−−=−∫∫|21|ln21)21(21121211.(4)∫−332xdx;解CxCxxdxxdx+−−=+−⋅−=−−−=−∫∫−3232313)32(21)32(2331)32()32(3132.(5)∫−dxeaxbx)(sin;解Cbeaxabxdebaxdaxadxeaxbxbxbx+−−=−=−∫∫∫cos1)()(sin1)(sin.(6)∫dtttsin;解∫∫+−==Cttdtdtttcos2sin2sin.(7)∫;⋅xdxx210sectan解∫⋅xdxx210sectanCxxxd+==∫1110tan111tantan.(8)∫xxxdxlnlnln;解Cxxdxxdxxxxxdx+===∫∫∫|lnln|lnlnlnlnln1lnlnlnln1lnlnln.(9)∫+⋅+dxxxx2211tan;解∫+⋅+dxxxx2211tan2222211cos1sin11tanxdxxxdx+++=++=∫∫Cxxdx++−=++−=∫|1cos|ln1cos1cos1222.(10)∫xxdxcossin;解Cxxdxdxxxxxdx+===∫∫∫|tan|lntantan1tanseccossin2.(11)∫−+dxeexx1;解∫−+dxeexx1Cedeedxeexxxxx+=+=+=∫∫arctan11122.(12)∫;−dxxex2解.21)(212222Cexdedxxexxx+−=−−=−−−∫∫(13)∫;⋅dxxx)cos(2解Cxxdxdxxx+==⋅∫∫)sin(21)()cos(21)cos(2222.(14)∫−dxxx232;解CxCxxdxdxxx+−−=+−−=−−−=−∫∫−2212221223231)32(31)32()32(6132.(15)∫−dxxx4313;解∫∫+−−=−−−=−Cxxdxdxxx|1|ln43)1(11431344443.(16)∫;++dttt))sin((cos2ϕωϕω解Cttdtdttt++−=++−=++∫∫)(cos31)cos()(cos1)sin()(cos322ϕωωϕωϕωωϕωϕω.(17)∫dxxx3cossin;解CxCxxxddxxx+=+=−=−−∫∫2233sec21cos21coscoscossin.(18)∫−+dxxxxx3cossincossin;解)sincos(cossin1cossincossin33xxdxxdxxxxx+−−=−+∫∫Cxxxxdxx+−=−−=∫−3231)cos(sin23)cos(sin)cos(sin.(19)∫−−dxxx2491;解dxxxdxxdxxx∫∫∫−−−=−−22249491491)49(49181)32()32(1121222xdxxdx−−+−=∫∫Cxx+−+=2494132arcsin21.(20)∫+dxxx239;解Cxxxdxxdxxdxxx++−=+−=+=+∫∫∫)]9ln(9[21)()991(21)(9219222222223.(21)∫−dxx1212;解∫∫∫+−−=+−=−dxxxdxxxdxx)121121(21)12)(12(11212∫∫++−−−=)12(121221)12(121221xdxxdxCxxCxx++−=++−−=|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Cxxxdxxdxxdx+−=−==∫∫∫3223sin31sinsin)sin1(sincoscos.(24)∫;+dtt)(cos2ϕω解Cttdttdtt+++=++=+∫∫)(2sin4121)](2cos1[21)(cos2ϕωωϕωϕω.(25)∫;xdxx3cos2sin解∫xdxx3cos2sinCxxdxxx++−=−=∫cos215cos101)sin5(sin21.(26)∫dxxx2coscos;解Cxxdxxxdxxx++=+=∫∫21sin23sin31)21cos23(cos212coscos.(27)∫;xdxx7sin5sin解Cxxdxxxxdxx++−=−−=∫∫2sin4112sin241)2cos12(cos217sin5sin.(28)∫;xdxxsectan3解xdxxdxxxxdxxsectantansectansectan223∫∫∫=⋅=Cxxxdx+−=−=∫secsec31sec)1(sec32.(29)∫−dxxx2arccos2110;解Cxd