<s>[INST] Find $\int\arcsin(\sqrt{x})dx$ 
Find $\displaystyle\int\arcsin(\sqrt{x})dx$

My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
 [/INST] Note that 
$$
d/dx((x - 1/2) \sin^{-1}(\sqrt x)) = \frac{x - 1/2}{2 \sqrt{x(1 - x)} } + \sin^{-1}(\sqrt x )
$$
So you can easily integrate
$$
\int\sin^{-1}\sqrt{x}dx \\
= (x - 1/2) \sin^{-1}(\sqrt x) - \int \frac{x - 1/2}{2 \sqrt{x(1 - x)} } dx \\
= (x - 1/2) \sin^{-1}(\sqrt x) + \frac{ \sqrt{x(1 - x)}}{2}
$$
</s>