Different ways to tackle the integral $\int_0^1\sqrt\frac x{1-x}\,dx$ $$\int_0^1\sqrt\frac x{1-x}\,dx$$
I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$.
I would like to see different approaches, can you provide them?

Here is another way that involves rationalising the numerator first.
For $0 \leqslant x < 1$ we can write
\begin{align*}
\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx &= \int_0^1 \sqrt{\frac{x}{1 - x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} \, dx\\
&= \int^1_0 \frac{x}{\sqrt{x - x^2}} \, dx
\end{align*}
Now rewriting the numerator as the derivative of the denominator we have 
\begin{align*}
\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx &= -\frac{1}{2} \int^1_0 \frac{(1 - 2x) - 1}{\sqrt{x - x^2}} \, dx\\
&= -\frac{1}{2} \int^1_0 \frac{1 - 2x}{\sqrt{x - x^2}} \, dx + \frac{1}{2} \int^1_0 \frac{dx}{\sqrt{x - x^2}} \, dx\\
&= I_1 + I_2
\end{align*}
The first of these integrals can be found using a substitution of $x = u + 1/2$. The result is 
$$I_1 = \int^{1/2}_{-1/2} \frac{u}{\sqrt{1/4 - u^2}} \, du = 0,$$
as the integrand is odd between symmetric limits.
The second integral can be found by first completing the square. As 
$$x - x^2 = \frac{1}{4} - \left (x - \frac{1}{2} \right )^2,$$
we have
$$I_2 = \frac{1}{2} \int^1_0 \frac{dx}{\sqrt{\frac{1}{2^2} - \left (x - \frac{1}{2} \right )^2}} \, dx = \frac{1}{2} \sin^{-1} (2x - 1) \Big{|}^1_0 = \frac{\pi}{2}.$$
Thus
$$\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx = \frac{\pi}{2}.$$
