求不定积分：dx/(1+tanx)

答案是:(x+ln(sinx+cosx))/2+c,但不知道解题的过程是怎样的.请教高手

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∫1/(1+tanx)dx
=∫1/(1+sinx/cosx)dx
=∫cosx/(cosx+sinx)dx
=∫cosx(cosx-sinx)/(cosx+sinx)(cosx-sinx)dx
=∫(cos²x-sinxcosx)/(cos²x-sin²x)dx
=[∫(1+cos2x-sin2x)/cos2xdx]/2
=[∫(1+cos2x-sin2x)/cos2xd2x]/4
=(∫sec2xd2x+∫d2x+∫tan2xd2x)/4
=ln|sec2x+tan2x|/4+x/2+ln|cos2x|/4+C
=x/2+ln|cos2x(sec2x+tan2x)|/4+C
=x/2+ln(1+sin2x)/4+C

你的姿蚂答案跟我的结果是一样的，只不迹链埋过继续作变形

x/2+ln(1+sin2x)/4
=x/2+ln(sin²x+2sinxcosx+cos²x)/4
=x/2+ln(sinx+cosx)²/4
=x/2+ln√(sinx+cosx)²/唤孙2
=[x+ln(sinx+cosx)]/2

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把dx/中竖(1+tanx)化成cosxdx/(cosx+sinx)=d(sinx)/根号2*sin(x+派/4)
=（1/根号2）d(sin(x+派/4))/sin(x+派/4)=（历培迅1/根号2）*
ln(sin(x+派/肢此4))+c

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