Evaluating $\int_{0}^{1}\sqrt{\frac{x}{1-x}}\mathrm{d}x$

Question

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

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?

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 '19 at 04:34
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 '19 at 04:44
If it is not a definite integral from outer space we have already covered one (almost) like it. — Jyrki Lahtonen, Aug 12 '19 at 04:46
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 '19 at 04:46
DanDan, that's ok. Experience helps in finding those. While the on-site search is bad for questions with complicated formulas, TeX and Approach0 often help. Check it out! — Jyrki Lahtonen, Aug 12 '19 at 04:48

score 0 · Accepted Answer · answered Aug 12 '19 at 04:38

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:)

Evaluating $\int_{0}^{1}\sqrt{\frac{x}{1-x}}\mathrm{d}x$

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