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I was wondering why the following is true:

$$\int_0^1 \frac 1 { \sqrt{ x (1 - x) } } \, \mathrm d x = \pi$$

It is easy to obtain this result by doing a trig substitution but it's messy and not very enlightening. What is the quickest or most elegant method? I am convinced there is a cleverer way if only because the integral evaluates to $\pi$ - there must be a nice reason for this.

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    $\begingroup$ The derivative of $\arcsin(2x-1)$ gives the integrand. $\endgroup$ – coffeemath Jan 30 '15 at 20:03
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Let $x=u^2$, $dx = 2 u \,du $. Then the integral is

$$2 \int_0^1 \frac{du}{\sqrt{1-u^2}} $$

Now let $u=\sin{t}$, $du = \cos{t} \, dt$. Then the integral is

$$2 \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}} = 2 \frac{\pi}{2} = \pi$$

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  • $\begingroup$ Ah, that was the substitution I was looking for. Thank you $\endgroup$ – user85798 Jan 30 '15 at 20:20
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    $\begingroup$ In one blow one can let $x=\sin^2 t$. $\endgroup$ – André Nicolas Jan 30 '15 at 20:23
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    $\begingroup$ @AndréNicolas: yes but I like to show my elemental substitutions in a problem like this. The pattern recognition requirements are lighter that way. $\endgroup$ – Ron Gordon Jan 30 '15 at 20:30
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    $\begingroup$ @RonGordon: Good point, a rational process is better than a magic substitution. $\endgroup$ – André Nicolas Jan 30 '15 at 20:34
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The integral equals $$ \mathrm{B}(1/2,1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{\sqrt \pi \sqrt \pi}{1} = \pi. $$

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