4-3

4-3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.xdx