To Leibniz

Differentiating the given integral with respect to a will produce a form of the integral asked for in terms of a. We use a=2 to get the desired answer.

$\displaystyle-\frac{d}{da}\bigg(\int_0^\pi\frac{dx}{a-\cos{x}}\bigg)\bigg|_{a=2}=\int_0^\pi\frac{dx}{(2-\cos{x})^2}$

$\displaystyle-\frac{d}{da}\int_0^\pi\frac{dx}{a-\cos{x}}=-\frac{d}{da}\frac{\pi}{\sqrt{a^2-1}}=\frac{a\pi}{(a^2-1)^\frac3{2}}\quad\text{And we use }a=2\text{ to get the asked form}$

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$\textbf{Proof of given:}$

Using a Weiestrass substitution: $\displaystyle\cos{x}=\frac{1-t^2}{1+t^2}\quad dx=\frac{2dt}{1+t^2}$

$\displaystyle\int_0^\pi\frac{dx}{a-\cos{x}}=\int_0^\infty\frac2{(a-1)+(a+1)t^2}dt= \frac\pi{\sqrt{a^2-1}}\quad\text{agreeing with the given}$