Question

根号下（1-x/1+x）的不定积分怎么求

Answer 1

令x=cost
则原式=∫√（1-cost）/(1+cost)dcost
=∫√（1-cos^2t）/(1+cost)^2dcost
∫-sin^2t/(1+cost)dt
=∫（cos^2t-1）/(1+cost)dt
=∫cost-1dt
=-sint-t+c
=-√(1-x^2) - arccost + c