求不定积分：分子1+sinx分母1+cosx

Question

如题所述

Answer 1

万能公式：t=tanx/2;
sinx=2t/(1+t^2)
cosx=(1-t^2)/(1+t^2)
1+sinx=(t+1)^2/(1+t^2)
1+cosx=2/(1+t^2)
∫(1+sinx)/(1+cosx)dx=∫(t+1)^2/(1+t^2)dt
这下划归为有理式问题。可积分。

Answer 2

这个用万能代换
令
sinx=2tan(x/2)/[1+tan^2(x/2)]
cosx=[1-tan^2(x/2)]/[1+tan^2(x/2)]
所以(1+sinx)/(1+cosx)
={1+2tan(x/2)/[1+tan^2(x/2)]}/{1+[1-tan^2(x/2)]/[1+tan^2(x/2)]}
=[2tan(x/2)+1+tan^2(x/2)]/[1-tan^2(x/2)+1+tan^2(x/2)]
=[2tan(x/2)+1+tan^2(x/2)]/2
=[2tan(x/2)+sec^2(x/2)]/2
=tan(x/2)+sec^2(x/2)/2
会积分了吧？本回答被网友采纳

Answer 3

=∫1/(1+cosx)dx+∫sinx/(1+cosx)dx
=∫1/2cos^2(x/2)dx-∫1/(1+cosx)d(cosx+1)
=∫[sec(x/2)]^2d0.5x-∫1/(1+cosx)d(cosx+1)
=tan(x/2)-ln|1+cosx|+C