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

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.

• The derivative of $\arcsin(2x-1)$ gives the integrand. – coffeemath Jan 30 '15 at 20:03

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

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

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

The integral equals $$\mathrm{B}(1/2,1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{\sqrt \pi \sqrt \pi}{1} = \pi.$$