integration - Different ways to tackle the integral $int_0^1sqrtfrac x{1-x},dx$

$$\int_0^1\sqrt\frac x{1-x}\,dx$$ 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?

Answer

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 $$\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx = \frac{\pi}{2}.$$