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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?

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    $\begingroup$ For starters, change variables to $u = x/a$. $\endgroup$ – eyeballfrog Oct 15 '18 at 19:52
  • $\begingroup$ Ok, I understand that proof $\int_0^1{\frac{\tan^{-1}{\left(\sqrt{\frac{2-u^2}{4-u^2}}\right)}}{\sqrt{4-u^2}}du} = \frac{\pi^2}{32}$ will imply my statment, anyway I can't solve this $\endgroup$ – Fabio Oct 15 '18 at 19:58
  • $\begingroup$ Integrating by parts gives $\int_0^1 \tan^{-1}\left(\sqrt{\frac{2-u^2}{4-u^2}}\right)/\sqrt{4-u^2} du = \pi^2/36 + \int_0^1 \frac{u\sin^{-1}(u/2)}{(3-u^2)\sqrt{(3-u^2)^2-1}}du$. Still difficult, but at least the nasty argument of the arc trig function is gone. $\endgroup$ – eyeballfrog Oct 15 '18 at 20:20
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    $\begingroup$ See math.stackexchange.com/q/2955676/44121 $\endgroup$ – Jack D'Aurizio Oct 15 '18 at 21:03
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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

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    $\begingroup$ Man i've just finished to solve this, halfway in the Coxeter solution you have to stop and change direction, this integral is absolutely crazy. Thanks for showing me the way $\endgroup$ – Fabio Oct 16 '18 at 21:48

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