求∫（arcsinx)^2dx=?

网友回答

令arcsinx=t x=sint
∫（arcsinx)^2dx
=∫t^2costdt
=∫t^2dsint
=t^2sint-2∫tsintdt
=t^2sint+2∫tdcost
=t^2sint+2(tcost-∫costdt)
=t^2sint+2(tcost-sint)
∫（arcsinx)^2dx=x(arcsinx)^2+2(arcsinxcos(arcsinx)-x)
======以下答案可供参考======
供参考答案1:
令x=sint,t∈（-Pi2,Pi/2]
则arcsinx=arcsin(sint)=t
∫（arcsinx)^2dx
=∫t^2dsint
=∫t^2(sint)'dt
=t^2*sint-∫2tsintdt
=t^2*sint+2∫t(cost)'dt
=t^2*sint+2tcost-2∫costdt
=t^2*sint+2tcost-2∫dcost
=t^2*sint+2tcost-2cost
中间用了两次分部积分法
然后再代回去
∫（arcsinx)^2dx
==(arcsinx)^2*x+2arcsinx*(1-x^2)^0.5-2*(1-x^2)^0.5