Trig. definite Integral

[jacks]

\(\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi \)

[Tolaso J Kos]

Good evening...

Well, this integral took me a long time to go around it... but with no success actually... until a friend of mine noticed that the integral is symmetric to \(\displaystyle \frac{\pi }{8} \).Meaning that from \(\displaystyle \frac{\pi }{8} \) to \(\displaystyle \frac{\pi }{4} \) is the negative of zero to \(\displaystyle \frac{\pi }{4} \), thus the integral equals zero..

Wow... !!

[Grigorios Kostakos]

Well, this integral took me a long time to go around it... but with no success actually... until a friend of mine noticed that the integral is symmetric to \(\displaystyle \frac{\pi }{8} \).Meaning that from \(\displaystyle \frac{\pi }{8} \) to \(\displaystyle \frac{\pi }{4} \) is the negative of zero to \(\displaystyle \frac{\pi }{4} \), thus the integral equals zero..

Apostolos

The only root of the function \(f(x)=\arctan\Bigl(\tfrac{\sqrt{2}\cos(3 x)}{(2\cos (2x)+ 3)\sqrt{\cos (2x)}}\Bigr)\) in the interval \(\bigl({0,\frac{\pi}{4}}\bigr)\) is \(\frac{\pi}{6}\) and \(\lim_{x\rightarrow{\frac{\pi}{4}}^{-}}f(x)=-\infty\).


So, maybe the given integral is really equal to zero, but I don't think that it is easy to prove.