Question

高等数学不定积分∫dx/[1+√(1-x²)] 用换元积分法怎么求?

Answer 1

令 x=sinu,则
I = ∫dx/[1+√(1-x²)] = ∫cosudu/(1+cosu)
= ∫[1-1/(1+cosu)]du = u - ∫[sec(u/2)]^2d(u/2)
= u - tan(u/2)+C
= arcsinx - [1-√(1-x²)] /x +C