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 ?
 A: Hint. Note that
$$-\int\frac{y^2}{\sqrt{1-y^2}}\,dy=\int\frac{1-y^2}{\sqrt{1-y^2}}\,dy-\int\frac{dy}{\sqrt{1-y^2}}=\int\sqrt{1-y^2}\,dy-\arcsin(y).$$
A: I would proceed as follows:\begin{align}\int\frac{y^2}{\sqrt{1-y^2}}\,\mathrm dy&=\int y\frac y{\sqrt{1-y^2}}\,\mathrm dy\\&=-y\sqrt{1-y^2}+\int\sqrt{1-y^2}\,\mathrm dy.\end{align}
A: $$\int\frac{-y^2}{\sqrt{1-y^2}}dy=\int\frac{1-y^2-1}{\sqrt{1-y^2}}dy
=\int\frac{1-y^2}{\sqrt{1-y^2}}dy-\int\frac{1}{\sqrt{1-y^2}}dy=$$ $$=
\int\sqrt{1-y^2}dy-\arcsin(y)=\left[y=\sin{t}, dy=\cos{t}\ dt\right]=\int\cos^2{t}dt-\arcsin(y)=$$ $$=\frac12\int(1+\cos{2t})dt-\arcsin(y)=\frac12\left[t+\frac12\sin{2t}\right]_{t=\arcsin{y}}-\arcsin(y)+C=$$ $$=\frac12\arcsin{y}+\frac14\sin{(2\arcsin{y})}-\arcsin(y)+C=$$ $$=-\frac12\arcsin{y}+\frac12\sin{(\arcsin{y})}\cos{(\arcsin{y})}+C=$$
$$=-\frac12\arcsin{y}+\frac12y\sqrt{1-y^2}+C$$
A: Hint. Notice that $\int\frac{1}{\sqrt{1-y^2}}\mathrm{d}y=\sin^{-1}(y)$, so you could try writing $y^2$ as $y^2-1+1$ and splitting into two integrals:
$$\int\frac{y^2}{\sqrt{1-y^2}}\mathrm{d}y=\int\left(\frac{y^2-1}{\sqrt{1-y^2}}+\frac{1}{\sqrt{1-y^2}}\right)\mathrm{d}y\\=\int\left (\frac{1}{\sqrt{1-y^2}}-\sqrt{1-y^2}\right )\mathrm{d}y=\int\frac{1}{\sqrt{1-y^2}} \mathrm{d}y-\int\sqrt{1-y^2} \mathrm{d}y$$
Solving $\int\sqrt{1-y^2} \mathrm{d}y$ can be made via trigonometric substitution, setting $y=\sin u$.
A: 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
$$
\int\sin^{-1}\sqrt{x}dx \\
= (x - 1/2) \sin^{-1}(\sqrt x) - \int \frac{x - 1/2}{2 \sqrt{x(1 - x)} } dx \\
= (x - 1/2) \sin^{-1}(\sqrt x) + \frac{ \sqrt{x(1 - x)}}{2}
$$
A: By parts directly:
$$\begin{cases}u=\arcsin\sqrt x,&u'=\frac1{2\sqrt x\sqrt{1-x}}\\{}\\
v'=1,&v=x\end{cases}\;\;\implies \int\arcsin\sqrt x\,dx=x\arcsin\sqrt x-\frac12\int\sqrt\frac x{1-x}\,dx$$
and now substitute 
$$\;u^2=\frac x{1-x}\implies x(-u^2-1)=-u^2\implies x=\frac{u^2}{u^2+1}=1-\frac1{1+u^2}\implies$$
$$ dx=\frac{2u}{(1+u^2)^2}\;$$
and from here your integrals equals
$$x\arcsin\sqrt x-\int\frac{u^2}{(1+u^2)^2}\,du=x\arcsin x-\frac12\left(\arctan u-\frac u{1+u^2}\right)$$
and now go back to $\;x\;$ and etc.
