Question

这个不定积分怎么求

Answer 1

Answer 2

∫[(1-cosx)/sin²x]dx
=∫[(1-cosx)/(1-cos²x)]dx
=∫[1/(1+cosx)]dx
=(1/2)∫[1/cos²(x/2)]dx
=∫sec²(x/2)d(x/2)
=tan(x/2)+C。