∫1/[1+(1-x^2)^(1/2)]dx,求解答过程

∫1/[1+(1-x^2)^(1/2)]dx
= ∫1/[1+cost]d(sint) (令x=sint)
=∫(cost)/(1+cost)dt
=∫dt-∫dt/(1+cost)
=t+ctgt-csct+C
=arcsinx+ctgarcsinx-1/x +C
=arcsinx + (1-x^2)^(1/2)/x +C
再问： 答案是arcsinx +x/[1+ (1-x^2)^(1/2)]+C ，后面反三角函数转换不懂，可以教教我吗？ =∫dt-∫dt/(1+cost) =t+ctgt-csct+C 这也不懂
再答： =∫dt-∫dt/(1+cost) =t-∫(1-cost)sintsintdt =t-∫csctcsctdt+∫d(sint)/sintsint =t+ctgt-1/sint +C =t+cost/sint-1/sint +C =arcsinx+(1-x^2)^(1/2)/x-1/x +C =arcsinx+{[(1-x^2)^(1/2)-1][(1-x^2)^(1/2)+1]}/{x[(1-x^2)^(1/2)+1]}+C =arcsinx-x/[1+ (1-x^2)^(1/2)]+C