The integral ${\int }_{\pi /6}^{\pi /4}\frac{dx}{\sin 2x\left({\tan }^{5}x+{\cot }^{5}x\right)}$equals 

$$\begin{gathered} \mathrm{I}={\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/4}\frac{\mathrm{dx}}{\sin 2\mathrm{x}\left({\tan }^5\mathrm{x}+{\cot }^5\mathrm{x}\right)} \\ ={\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/4} \frac{\mathrm{dx}}{{\displaystyle \frac{2\tan \mathrm{x}}{1+{\tan }^2\mathrm{x}}} \left({\tan }^5\mathrm{x}+{\displaystyle \frac{1}{{\tan }^5\mathrm{x}}}\right)}\mathrm{dx} \\ ={\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/4} \frac{\sec 2\mathrm{x} \mathrm{dx}}{2\tan \mathrm{x} \left({\tan }^5\mathrm{x}+{\displaystyle \frac{1}{{\tan }^5\mathrm{x}}}\right)} \\ \mathrm{Put} \tan \mathrm{x}=\mathrm{t} \\ {\sec }^2 \mathrm{dx}=\mathrm{dt} \\ ={\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.}^1 \frac{\mathrm{dt}}{2\mathrm{t}\left({\mathrm{t}}^5+{\displaystyle \frac{1}{{\mathrm{t}}^5}}\right)} \\ =\frac{1}{2} {\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.}^{1 } \frac{{\mathrm{t}}^4 \mathrm{dt}}{\left({\mathrm{t}}^{10}+1\right)} \\ =\frac{1}{10} {\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.}^1 \frac{5{\mathrm{t}}^4 \mathrm{dt}}{{\left({\mathrm{t}}^5\right)}^2+1}=\frac{1}{10} {{\tan }^{-1}\left({\mathrm{t}}^5\right) }_{\frac{1}{\sqrt{3}}}^1=\frac{1}{10} \left[\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.-{\tan }^{-1} \frac{1}{9\sqrt{3}}\right] \end{gathered}$$