1/(1+cosx)的不定积分是怎么算啊 ∫dx/1+cosx=?

Question

如题所述

Answer 1

1+cosx=2[cos(x/2)]^2
1/(1+cosx)=0.5[sec(x/2)]^2
∫dx/(1+cosx)
=∫0.5[sec(x/2)]^2dx
=∫[sec(x/2)]^2d0.5x
=∫dtan(x/2)
=tan(x/2)+c