Evaluate $\int_{0}^\frac{1}{2}\frac{\sin^{-1}x}{\sqrt{1-x^2}} dx $ $$\int_0^\frac{1}{2} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx $$

My question for this is can I take out the square root so it would be 

$$\frac{1}{\sqrt{1-x^2}} \int_{0}^\frac{1}{2}{\sin^{-1}x} \,dx $$

So all i have to do is find the anti-derivative  of $\sin^{-1}x$ then multiply the $\dfrac{1}{\sqrt{1-x^2}}$ back in afterwards? Is there a simpler way?
Hint: let $u = \arcsin x$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$. You're not allowed to move the $\frac{1}{\sqrt{1-x^2}}$ term out in front since it has an $x$ in it and the variable of integration is $x$--so the integral depends on that term too!
I may suggest that this integral must be like this Integral arc sin x d(arc sin x) from 0 to 1/2. thus the answer must be π^2/72