# Evaluating $\int_{0}^{1}\sqrt{\frac{x}{1-x}}\mathrm{d}x$ [duplicate]

This question already has an answer here:

$$\int_0^{1}\sqrt{\frac{x}{1-x}}\mathrm{d}x=\frac{\pi}{2}$$

This integral seems to be an identity, since the antiderivative for $$\sqrt{\frac{x}{1-x}}$$ is somewhat cumbersome and the integrand has a vertical asymptote at $$x=1$$.

How do we evaluate this integral without resorting to a lookup table?

## marked as duplicate by Jyrki Lahtonen, Community♦Aug 12 at 4:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• $x=\sin^2(u)$ seems to work nicely. – robjohn Aug 12 at 4:33
• What makes you say that the antiderivative is particularly cumbersome? It's a rather straightforward computation, just let $t=\sqrt{\frac{x}{1-x}}$ for example. – Hans Lundmark Aug 12 at 4:34
• The antiderivative as given by Wolfram|Alpha is pretty long in my opinion. Also, could you please clarify on what you mean by $t=\sqrt{\frac{x}{1-x}}$? Thanks. – DanDan0101 Aug 12 at 4:44
• If it is not a definite integral from outer space we have already covered one (almost) like it. – Jyrki Lahtonen Aug 12 at 4:46
• Sorry, I think it is. I tried searching for the integral before I posted but couldn't find that post. Thanks for pointing this out. – DanDan0101 Aug 12 at 4:46

## 1 Answer

Put $$x=\sin^2t$$ $$\Longrightarrow dx=\sin 2t dt$$ $$\Longrightarrow \int_0^1\sqrt{\frac{x}{1-x}}dx=\int_0^{\frac{π}{2}}\tan t \sin 2t dt$$ $$=\int_0^{\frac{π}{2}}2\sin^2 t dt$$ $$=2×\frac{π}{4}=\frac{π}{2}$$ where the last step uses Walli's formula. Hope it helps:)