∫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
你的答案跟我的結(jié)果是一樣的,只不過繼續(xù)作變形
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)²/2
=[x+ln(sinx+cosx)]/2