4-2高等数学同济大学第六版本

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a1d(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,,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((cos2;解Cttdtdttt)(cos31)cos()(cos1)sin()(cos322.(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)(cos2;解Cttdttdtt)(2sin4121)](2cos1[21)(cos2.(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Cxdxddxxxxxx10ln210)arccos2(1021arccos10110arccos2arccos2arccos22arccos2.(30)dxxxx)1(arctan;解Cxxdxxdxxdxxxx2)(arctanarctanarctan2)1(arctan2)1(arctan.(31)221)(arcsinxxdx;解Cxxdxxxdxarcsin1arcsin)(arcsin11)(arcsin222.(32)dxxxx2)ln(ln1;解Cxxxxdxxdxxxxln1)ln()ln(1)ln(ln122.(33)dxxxxsincostanln;解xdxxxdxxxdxxxxtantantanlnsectantanlnsincostanln2Cxxdx2)tan(ln21tanlntanln.(34)dxxax222(a0);解dttadttatdtatatataxdxxax22cos1sincoscossinsin22222222令,CxaxaxaCtata222222arcsin22sin421.(35)12xxdx;解CxCtdttdtttttxxxdx1arccostansectansec1sec12令.或Cxxdxdxxxxxdx1arccos111111112222.(36)32)1(xdx;解Cttdttdttxxdxsincostan)1(tan1tan)1(3232令Cxx12.(37)dxxx92;解tdttdtttxdxxx222tan3)sec3(sec39sec9sec39令CxxCttdtt3arccos393tan3)1cos1(322.(38)xdx21;解CxxCttdtttdtttxxdx)21ln(2)1ln()111(11221令.(39)211xdx;解dttdtttdtttxxdx)2sec211()cos111(coscos11sin1122令CxxxCtttCtt211arcsincos1sin2tan.(40)21xxdx.解dttttttttdttttxxxdxcossinsincossincos21coscossin1sin12令Ctttttdttdt|cossin|ln2121)cos(sincossin12121Cxxx|1|ln21arcsin212.