Question

求不定积分 cotx/(1+sinx)dx要详细过程谢了

Answer 1

简单分析一下，答案如图所示

Answer 2

令u=sinx,则du=cosxdx,
∫cotx/(1+sinx) dx
=∫1/[u(u+1)] du
=∫[1/u-1/(u+1)]du
=∫1/u du-∫1/(u+1)d(u+1)
=ln|u|-ln|u+1|+C
=ln|u/(u+1)|+C
=ln|sinx/(sinx+1)|+C