Problem
Evaluate $$ \int_0^{\frac{\pi}{4}}\arctan\frac{\sqrt{2}\cos 3x}{(3+2\cos 2x)\sqrt{\cos 2x}}{\rm d}x.$$
The integrand is complicated. AlphaWolfram outputs the result is zero. If we consider making a subsitution, let $x=\dfrac{\pi}{4}-u$, then $$\int_0^{\frac{\pi}{4}}\arctan\frac{\sqrt{2}\cos 3x}{(3+2\cos 2x)\sqrt{\cos 2x}}{\rm d}x=\int_0^{\frac{\pi}{4}}\arctan\frac{\sqrt{2}\cos\left(3u+\dfrac{\pi}{4}\right)}{(3+2\sin 2u)\sqrt{\sin 2x}}{\rm d}u,$$ which seems to be helpless.
