I conjecture that $$\small \int_0^{\pi/2} \frac{\cos ^2(x) \left(-2 \log \left(4^{-\sin ^2(x)} \sin ^{-4 \sin ^2(x)}(x)\right)-4 \log (\cos (x))+\cos (2 x) (4 \log (\cos (x))+\pi +\log (4))\right)}{\cos (4 x)+3} \, dx$$ $$=\frac{\pi ^2}{8}-\frac{\pi ^2}{8 \sqrt{2}}+\frac{\pi \log (2)}{4 \sqrt{2}}$$

Is there a simple way to calculate it? If dividing the integrand, we get a tough nut ...

EDIT: It seems we may rewrite it as

$$\frac{1}{2}\int_0^{\pi/2} \frac{\pi+4 \tan ^2(x) \log (\tan (x))}{1+ \tan ^4(x)} \, dx-\frac{\pi (\pi -2 \log (2))}{8 \sqrt{2}}$$

Using Khallil's idea of variable change in the last integral, we complete the solution.


  • 1
    $\begingroup$ Just a thought, but have you considered $x \mapsto \pi/2 - x$? $\endgroup$ – Kari Jul 27 '15 at 21:43
  • $\begingroup$ @Khallil I'll check that and see if it leads somewhere. $\endgroup$ – user 1357113 Jul 27 '15 at 21:45
  • 1
    $\begingroup$ The answer is so neat! $\endgroup$ – xpaul Jul 28 '15 at 14:07

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