$I = \int^{\pi/4}_{\pi/6} \frac{dx}{\sin2x\left(\tan^{5} x + \cot^{5} x\right)} $
$ I = \frac{1}{2} \int^{\pi/4}_{\pi/6} \frac{\tan^{4} x \sec^{2} x dx}{\left(1+\tan^{10} x\right)} $ Put $ \tan^{5} x = t $
$ I = \frac{1}{10} \int^{1}_{\left(\frac{1}{\sqrt{3}}\right)^{5}} \frac{dt}{1+t^{2}} = \frac{1}{10} \left(\frac{\pi}{4} -\tan^{-1} \frac{1}{9\sqrt{3}}\right) $