Using integration by parts, we set:

`u=arcsin\ x`, giving `du=1/sqrt(1-x^2)dx`.

Then `dv=dx` and integrating gives us `v=x`.

We now use:

`intu\ dv=uv-intv\ du`

This gives us:

`int arcsin\ x\ dx` `=x\ arcsin\ x-intx/(sqrt(1-x^2))dx`

Now, for that remaining integral, we just use a substitution (I'll use `p` for the substitution since we are using `u` in this question already):

`p = 1 − x^2`

So `dp=-2x\ dx`

This will yield:

`intx/(sqrt(1-x^2))dx =-1/2int(dp)/sqrtp`



So our final answer is:

`int arcsin\ x\ dx =x\ arcsin\ x-(-sqrt(1-x^2))+K `

`= \x\ arcsin\ x+sqrt(1-x^2)+K`

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