求(1+sinx)\(1+cosx)与e^xdx的积的不定积分
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求(1+sinx)\(1+cosx)与e^xdx的积的不定积分
![求(1+sinx)\(1+cosx)与e^xdx的积的不定积分](/uploads/image/z/4147600-40-0.jpg?t=%E6%B1%82%EF%BC%881%2Bsinx%29%5C%281%2Bcosx%29%E4%B8%8Ee%5Exdx%E7%9A%84%E7%A7%AF%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86)
1/(1+cosx)=1/[1+2cos²(x/2)-1]
=sec²(x/2)/2
所以原式=∫e^xsec²(x/2)/2dx+∫e^xsinx*sec²(x/2)/2dx
=∫e^xsec²(x/2)d(x/2)+∫e^x2sin(x/2)cos(x/2)*sec²(x/2)/2dx
=∫e^xdtan(x/2)+∫e^x*sin(x/2)/cos(x/2)dx
=∫e^xdtan(x/2)+∫tan(x/2)de^x
=e^x*tan(x/2)-∫tan(x/2)de^x+∫tan(x/2)de^x
=e^x*tan(x/2)+C
=sec²(x/2)/2
所以原式=∫e^xsec²(x/2)/2dx+∫e^xsinx*sec²(x/2)/2dx
=∫e^xsec²(x/2)d(x/2)+∫e^x2sin(x/2)cos(x/2)*sec²(x/2)/2dx
=∫e^xdtan(x/2)+∫e^x*sin(x/2)/cos(x/2)dx
=∫e^xdtan(x/2)+∫tan(x/2)de^x
=e^x*tan(x/2)-∫tan(x/2)de^x+∫tan(x/2)de^x
=e^x*tan(x/2)+C
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