$\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$

$\begin{array}{l}

=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{5} x \cdot \sec ^{2} x}{2 \frac{\sin x}{\cos x}\left(\left(\tan ^{5} x\right)^{2}+1\right)} \\

=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4} \tan ^{4} x \cdot \sec ^{2} x}{\left(\tan ^{5} x\right)^{2}+1} d x

\end{array}$

Let $\tan ^{4} x=t$

$5 \tan ^{4} x \cdot \sec ^{2} x d x=d t$

When $x \rightarrow \frac{\pi}{4}$ then $t \rightarrow 1$

and $x \rightarrow \frac{\pi}{6}$ then $t \rightarrow\left(\frac{1}{\sqrt{3}}\right)^{5}$

$\begin{aligned}

\therefore \quad & \mathrm{I}=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{d t}{t^{2}+1} \\

&=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)

\end{aligned}$