Proof of this integral identity $\int_0^a{\frac{\tan^{-1}{\left(\sqrt{\frac{2a^2-x^2}{4a^2-x^2}}\right)}}{\sqrt{4a^2-x^2}}dx} = \frac{\pi^2}{32}$

Question

I would like to proof this identity I conjectured

$\forall a\in \mathbb{R}^+$ $$\int_0^a{\frac{\tan^{-1}{\left(\sqrt{\frac{2a^2-x^2}{4a^2-x^2}}\right)}}{\sqrt{4a^2-x^2}}dx} = \frac{\pi^2}{32}$$

My first tought was to evaluate this integral in a special case and then generalized it. Sadly I'm unable to evaluate this integral with any kind of technique I know (various substitution or integration by parts), for any value of $a$, moreover this is my first proof which includes an integral.

I don't know what kind of tools or theorems i'm supposed to use in this kind of proofs, can you give me some hint or show me a path to solve this kind of problems?

Answer 1

This is known as a Coxeter integral:$$ I=\int_0^1{\frac{\tan^{-1}{\left(\sqrt{\frac{2-u^2}{4-u^2}}\right)}}{\sqrt{4-u^2}}du}$$substitute $u=2\sin x$ $$I=\int_0^\frac{\pi}{6}\arctan \left(\sqrt{1-\frac12\sec^2 x}\right)dx=\int_0^\frac{\pi}{6}\arctan \left(\sqrt{\frac{\cos(2x)}{2\cos^2x}}\right)dx$$Now proceed as shown here: http://sos440.tistory.com/212