The integral \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)}\) is equal to

\(\text {Let } I =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)} \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\mathrm{~d} x}{2 \sin x \cos x\left(\tan ^5 x+\frac{1}{\tan ^5 x}\right)} \)
\( =\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec ^2 x}{\tan x\left(\frac{\tan ^{10} x+1}{\tan ^5 x}\right)} \mathrm{d} x \)
\( =\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^4 x \sec ^2 x}{\tan ^{10} x+1} \mathrm{~d} x\)
Put \(\tan ^5 x=\mathrm{t} \Rightarrow 5 \tan ^4 x \sec ^2 x \mathrm{~d} x=\mathrm{dt}\)
\(\therefore \mathrm{I} =\frac{1}{2} \int_{\frac{1}{9 \sqrt{3}}}^1 \frac{\frac{\mathrm{dt}}{5}}{\mathrm{t}^2+1} \)
\( =\frac{1}{10}\left[\tan ^{-1} \mathrm{t}\right]_{\frac{1}{9 \sqrt{3}}}^1 \)
\( =\frac{1}{10}\left[\tan ^{-1} 1-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right] \)
\( =\frac{1}{10}\left[\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right]\)