how to integrate (arcsinx)^2 dx?

Question

can you help me confirm my answer

Answer 1

Yes, you can use parts, but why not try
u = arcsin x, x = sin u, dx = cosu du
cos u = √(1-x²)
So you get
∫u² cos u du,
which is an easy integration by parts. Let's do it by
the "tic-tac-toe" method:
u² cos u
2u sin u
2 -cos u
0 - sin u
So ∫ u² cos u = (u²-2)sin u + 2u cos u
Now let's substitute back to get your original answer. We get
∫ (arcsin x)² = x arcsin² x -2x + 2*√(1-x²) arcsin x + C,
or, rearranging,
x arcsin² x + 2√(1-x²) arcsin x - 2x + C
This matches the answer in my integral tables
and that of the Wolfram integrator, so it's correct.

Answer 2

try integration by parts with u=arcsinx and dv=dx.