Given the density function, show that $c sin^{-1}(\sqrt{x})$ is the distribution function and find c. 
I was able to find c which is 
$F(x) = c sin^{-1}(\sqrt{x})$ if $0 < x < 1$
$F(1) - F(0) $
$c (sin^{-1}(\sqrt{1})-sin^{-1}(\sqrt{0})) = 1$ 
$c (\frac{\pi}{2} - 0) = 1$ 
$c (\frac{\pi}{2}) = 1$ 
$c = \frac{1}{(\frac{\pi}{2}) }$ 
$c = \frac{2}{\pi}$ 
However, I am having trouble with the integration. My attempt would be to treat the $\frac{1}{\pi}$ as a constant and have this 
$\frac{1}{\pi} \int_{0}^{1} \frac{1}{\sqrt{x(1-x)}} dx$
Then, I'm not so sure. Would I have to use integration by parts?
I would let $ u = sin^{-1}(\sqrt{x})$, $du = \frac{1}{\sqrt{x(1-x)}}$, $dv= \frac{1}{\pi}$ and $v =\frac{1}{\pi} x$. 
$udv = uv - \int vdu$
$  sin^{-1}(\sqrt{x}) \cdot \frac{1}{\pi} = sin^{-1}(\sqrt{x}) \cdot \frac{1}{\pi} x - \int \frac{1}{\pi}x \cdot \frac{1}{\sqrt{x(1-x)}}  $
I'm not sure if I'm on the right track.
 A: The name $u$-substitution really doesn't mean anything, since it could mean too many things.
Observe the indefinite integral and notice that it would be nice if the $\frac1{\sqrt{x} }$ can be "canceled" with $dx$
$$\mathcal{I} = \frac1{\pi} \int \frac1{ \sqrt{1 - x} } \cdot \frac1{ \sqrt{x} } dx$$
The idea roughly speaking is that since $x$ is the square of $\sqrt{x}$ as in: let $x = u^2$ where $u = \sqrt{x}$ such that $dx = 2u\,du$ and the indefinite integral becomes
$$\mathcal{I} = \frac1{\pi} \int \frac1{ \sqrt{1 - u^2} } \cdot \frac1u \,2u \, du = \frac2{\pi} \int \frac1{ \sqrt{1 - u^2} } du$$
One sees the $1 - u^2$ and recalls something of the same form: $1 - \sin^2\theta = \cos^2\theta~$. Thus the next natural substitution is $u = \sin\theta$, so that $du = \cos\theta\,d\theta$ and
$$\mathcal{I} = \frac2{\pi} \int \frac1{ |\cos\theta| } \, \cos\theta \,d\theta$$
where the composite substitution is $x = u^2 = \sin^2\theta$ or equivalently $\theta = \sin^{-1}(\sqrt{x})$.
Now, one goes back to the definite integral and check the upper and lower limits of integration to make sure that $0 < \theta < \frac{ \pi }2$ which means $|\cos\theta| = \cos\theta$ and everything will work out fine.
Note that the given density already has the normalization constant taken care of (the definite integral $\int_0^1 f(x) dx = 1$ as needed).
