Question

How do you integrate `int sin^-1x` by integration by parts method?

Answer 1

`xarcsinx+sqrt(1-x^2)+C`

Explanation:

After choosing `u=arcsinx` and `dv=dx`, `du=(dx)/sqrt(1-x^2)` and `v=x`

Hence,

`int arcsinx*dx=xarcsinx-int x*(dx)/sqrt(1-x^2)`

=`xarcsinx-int (xdx)/sqrt(1-x^2)`

=`xarcsinx+sqrt(1-x^2)+C`