# Evaluate $\int_0^{1}{\sqrt{\frac{x}{1-x}} dx}$ [closed]

How to evaluate the following integral? $$\int_0^{1}{\sqrt{\frac{x}{1-x}} dx}$$

I know the answer is $\pi /2$, but how?

It was mentioned somewhere that it can be solved using $x = \sin^2 \theta$, but I don't know how.

## closed as off-topic by Namaste, Siong Thye Goh, JMP, Claude Leibovici, user1551Sep 7 '17 at 7:33

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• do you mean $$\int\sqrt{\frac{x}{1-x}}dx$$? – Dr. Sonnhard Graubner Aug 26 '17 at 18:27
• Yes. I forgot to write it with sqrt. – titansarus Aug 26 '17 at 18:28

Let $x = \sin^2(\theta) \implies dx=2\sin(\theta)\cos(\theta) d\theta$. Consequently, \begin{align} \int_{0}^{1} \sqrt{\frac{x}{1-x}} dx & = 2 \int_{0}^{\pi/2} \frac{\sin(\theta)}{\cos(\theta)}\sin(\theta)\cos(\theta) d\theta\\ & = 2\int_{0}^{\pi/2} \sin^2(\theta)d\theta\\ & = \int_{0}^{\pi/2}(1-\cos(2\theta)) d\theta \\ & = \frac{\pi}{2}. \end{align}
• Isn't $1-sin^2 \theta = cos^2 \theta$? so how $sqrt{sin^2 \theta / cos^2 \theta}$ equals $2sin^2 \theta$ – titansarus Aug 26 '17 at 18:36