Evaluate:
$$\int \arcsin \left(\sqrt{x}\right)dx$$
I am really stuck here. I know that I have to use $u-$substitution and set $u = \sqrt{x}$, to get $\left(2\int \arcsin \left(u\right)\cdot \:udu \right)$ but how can I continue from here? Could I use integration by parts and use $u = \arcsin(u), v'=u$?
Best Answer
Substitute $y=\arcsin\sqrt x$. Then, $dx= \sin 2y \>dy$ and $$\int \arcsin \sqrt{x}\>dx =\int y \sin 2y \>dy \overset {IBP}= -\frac12 y \cos 2y +\frac14\sin2y+C $$