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习题43求下列不定积分:1.xdxxsin;解Cxxxxdxxxxxdxdxxsincoscoscoscossin.2.xdxln;解Cxxxdxxxxxdxxxdxlnlnlnlnln.3.xdxarcsin;解xxdxxxdxarcsinarcsinarcsindxxxxx21arcsinCxxx21arcsin.4.dxxex;解dxexexdedxxexxxxCxeCexexxx)1(.5.xdxxln2;解xdxxxxdxxdxxln31ln31ln31ln3332Cxxxdxxxx332391ln3131ln31.6.xdxexcos;解因为xdxexexdexexdexdxexxxxxxsinsinsinsinsincosxxxxxxdexexexdexecoscossincossinxdxexexexxxcoscossin,所以CxxeCxexexdxexxxx)cos(sin21)cossin(21cos.7.dxxex2sin2;解因为xxxxdexxexdedxxe22222cos22cos22cos22sin2sin82cos22cos42cos22222xdexedxxexexxxxxxxdexxexe2222sin82sin82cos2dxxexexexxx2sin162sin82cos2222,所以Cxxedxxexx)2sin42(cos1722sin22.8.dxxx2cos;解Cxxxdxxxxxxddxxx2cos42sin22sin22sin22sin22cos.9.xdxxarctan2;解dxxxxxxdxxdxx233321131arctan31arctan31arctan2232223)111(61arctan31161arctan31dxxxxdxxxxxCxxxx)1ln(6161arctan31223.10.xdxx2tan解xxdxxdxxdxxdxxxxdxxtan21sec)1(sectan2222Cxxxxxdxxxx|cos|lntan21tantan2122.11.xdxxcos2;解xxdxxxdxxxxxdxxdxxcos2sin2sinsinsincos2222Cxxxxxxdxxxxxsin2cos2sincos2cos2sin22.12.dttet2;解dtetetdedttetttt2222212121CteCetettt)21(214121222.13.xdx2ln;解xdxxxdxxxxxxxdxln2ln1ln2lnln222Cxxxxxdxxxxxxx2ln2ln12ln2ln22.14.xdxxxcossin;解xdxxxxxdxdxxxdxxx2cos412cos412cos412sin21cossinCxxx2sin812cos41.15.dxxx2cos22;解xdxxxxxxdxxdxxxdxxxsinsin2161sin2161)cos1(212cos2323222xdxxxxxxxxdxxxcoscossin2161cossin21612323Cxxxxxxsincossin216123.16.dxxx)1ln(;解dxxxxxdxxdxxx1121)1ln(21)1ln(21)1ln(222dxxxxx)111(21)1ln(212Cxxxxx)1ln(212141)1ln(2122.17.xdxx2sin)1(2;解xdxxxxxdxxdxx22cos212cos)1(212cos)1(212sin)1(222xxdxx2sin212cos)1(212xdxxxxx2sin212sin212cos)1(212Cxxxxx2cos412sin212cos)1(212.18.dxxx23ln;解xdxxxxxdxxxxxddxxx22333323ln13ln1ln1ln11lnlnxdxxxxxxxdxx22323ln13ln3ln11ln3ln1xxdxxxxdxxxxxxx1ln6ln3ln1ln16ln3ln123223dxxxxxxxx22316ln6ln3ln1Cxxxxxxx6ln6ln3ln123.19.dxex3;解ttxdetdtettxdxe223333令tttttdeetdtteet636322dteteetttt6632Ceteetttt6632Cxxex)22(33323.20.xdxlncos;解因为dxxxxxxxdx1lnsinlncoslncosdxxxxxxxxxdxxx1lncoslnsinlncoslnsinlncosxdxxxxxlncoslnsinlncos,所以Cxxxxdx)lnsinln(cos2lncos.21.dxx2)(arcsin;解dxxxxxxdxx22211arcsin2)(arcsin)(arcsin221arcsin2)(arcsinxxdxxdxxxxx2arcsin12)(arcsin22Cxxxxx2arcsin12)(arcsin22.22.xdxex2sin.解xdxeedxxexdxexxxx2cos2121)2cos1(21sin2,而dxxexexdexdxexxxx2sin22cos2cos2cosxdxexexedexxexxxxx2cos42sin22cos2sin22cos,Cxxexdxexx)2sin22(cos512cos,所以Cxxeexdxexxx)2sin22(cos10121sin2.