Compute $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$ I haven't integrated in a while... so I'm a bit rusty. I have to integrate $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$. Should I apply some change of variables? If I take $u=x-1/2$, I get something of the form $\int\frac{1}{\sqrt{u^2+1/4}}du$ (along with another integral, which I can integrate), but I don't know how to deal with this one.
 A: Using $x=\sin^2 \theta$, then
\begin{align}
\int^1_0 \sqrt{\frac{x}{1-x}}\ dx =&\ \int^\frac{\pi}{2}_0\sqrt{\frac{\sin^2\theta}{1-\sin^2\theta}} d(\sin^2\theta)\\
=&\ \int^{\pi/2}_0 \tan\theta\ d(\sin^2\theta)\\
=&\ 2\int^{\pi/2}_0 \sin^2\theta\ d\theta = \frac{\pi}{2}
\end{align}
A: Apply another change of variables to get
$$\int\frac1{\sqrt{v^2+1}}dv,$$
and from there, try $v = \sinh x$. Use the following identities:
$$\frac{d}{dx}\sinh x = \cosh x\\
\cosh^2x - \sinh x^2 = 1$$
Everything should cancel out nicely; you'll end up with just $\int dx$, which is to say $\sinh^{-1}v$.
A: Use the property $$ \int_0^a f(x) ~dx = \int_0^a f(a-x) ~dx $$
$$ \mathscr{I} =\int_0^1\frac{x}{\sqrt{x(1-x)}}~ dx ~= ~\int_0^1\frac{1-x}{\sqrt{(1-x)x}}~ dx $$
Add both the integrals
$$ 2 \mathscr{I} = \int_0^1\frac{x+1-x}{\sqrt{x(1-x)}}~ dx ~= \int_0^1\frac{1}{\sqrt{x(1-x)}}~ dx $$
Now, this integral can be easily evaluated by completing the square
$$ \mathscr{I} = \dfrac12 \int_0^1 \frac{1}{\sqrt{\frac14-(x-\frac12)^2}}~ dx  $$
$$ \mathscr{I} = \dfrac12 ~\sin^{-1} \left( \dfrac{x-\frac12}{\frac12}  \right)_0^1 = \color{red}{\dfrac{\pi}{2}} $$
