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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?

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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.

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    $\begingroup$ $x=\sin^2(u)$ seems to work nicely. $\endgroup$ – robjohn Aug 12 at 4:33
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    $\begingroup$ 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. $\endgroup$ – Hans Lundmark Aug 12 at 4:34
  • $\begingroup$ 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. $\endgroup$ – DanDan0101 Aug 12 at 4:44
  • $\begingroup$ If it is not a definite integral from outer space we have already covered one (almost) like it. $\endgroup$ – Jyrki Lahtonen Aug 12 at 4:46
  • $\begingroup$ 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. $\endgroup$ – DanDan0101 Aug 12 at 4:46
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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:)

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