∫sinx /(1+cosx) dx = ∫tan(x/2) dx
1+cosx = 2 cos²(x/2)，[tan(x/2)] ' = (1/2) sec²(x/2)
∫x /(1+cosx) dx = ∫x d[tan(x/2)] = x * tan(x/2) - ∫tan(x/2) dx
原式 = x * tan(x/2) + C
∫(x+sinx)/(1+cosx)dx
=∫xdx/(1+cosx)+∫sinxdx/(1+cosx)
=∫xd(x/2)/[cos(x/2)]^2 +∫tan(x/2)dx
=∫xdtan(x/2)+∫tan(x/2)dx
=xtan(x/2)-∫tan(x/2)dx+∫tan(x/2)dx+C
=xtan(x/2) +C
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024