Showing that $ \int_{0}^{\pi / 4} \arctan \! \left( \sqrt{\frac{\cos 2x}{2 \cos^{2} x}} \right) \mathrm{d}{x} = \frac{\pi^{2}}{24} $.

Question

I was wondering if an expert in integration could kindly solve the following problem, which was posed in a mathematics competition (I can’t remember which one) and was unsolved by any participant.

Problem. Show that $$ \int_{0}^{\pi / 4} \arctan \! \left( \sqrt{\frac{\cos 2 x}{2 \cos^{2} x}} \right) \mathrm{d}{x} = \frac{\pi^{2}}{24}. $$

Thanks!

Answer 1

Considering the integral

$$ I(a) = \int_{0}^{\pi/4} \arctan\left(a \,\sqrt{\,\cos\left(\,2x\,\right) \over 2 \cos^{2}\left(\,x\,\right)\,}\,\,\right) \,{\rm d}x $$

and differentiating with respect to $a$ and making the substitution $\cos(x)^2=u$ gives rise to the integral

$$ I'(a)= \frac{\sqrt {2}}{2}\int _{1/2}^{1}\!{\frac {\sqrt {2\,u-1}}{\sqrt {-u+1} \left( 2\,{a}^{2}u-{a}^{2}+2\,u \right) }}{du} = \frac{\pi}{2}\,{\frac { \left( \sqrt {{a}^{2}+2}-1 \right) }{\sqrt {{a}^{2}+ 2}\left( {a}^{2}+1 \right)}}.$$

Integrating the above with respect to $a$ on the interval $[0,1]$ gives the desired result

$$ I(1)=\frac{\pi^2}{24}. $$

You need to verify the evaluation of the above integrals.