The value of \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)} \mathrm{dx}\) is

\( I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)} d x \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\frac{2 \tan x}{1+\tan ^2 x}\left(\tan ^5 x+\frac{1}{\tan ^5 x}\right)} d x \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\left(1+\tan ^2 x\right)}{2 \tan x\left(\tan ^5 x+\frac{1}{\tan ^5 x}\right)} d x\)
\(I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec ^2 x}{2 \tan x\left(\tan ^5 x+\frac{1}{\tan ^5 x}\right)} d x\)
Let \(\tan x=\mathrm{t}\) \(\sec ^2 x \mathrm{~d} x=\mathrm{dt}\) when \(x=\frac{\pi}{6}, \mathrm{t}=\frac{1}{\sqrt{3}}\) when \(x=\frac{\pi}{4}, t=1\)
\(\therefore I=\int_{\frac{1}{\sqrt{3}}}^1 \frac{d t}{2 t\left(t^5+\frac{1}{t^5}\right)} \)
\( I=\int_{\frac{1}{\sqrt{3}}}^1 \frac{t^4}{2\left(t^{10}+1\right)} d t \)
\( \text { Let } t^5=u \)
\( \therefore 5 t^4 d t=d u\)
When \(\mathrm{t}=\frac{1}{\sqrt{3}}, \mathrm{u}=3^{\frac{-5}{2}}\)
When \(\mathrm{t}=1, \mathrm{u}=1\)
\(\therefore I =\frac{1}{10} \int_{\frac{-5}{32}}^1 \frac{d u}{\left(u^2+1\right)} \)
\( =\frac{1}{10}\left(\tan ^{-1} u\right)_{\frac{-5}{2}}^1 \)
\( =\frac{1}{10}\left[\tan ^{-1} 1-\tan ^{-1}\left(3^{\frac{-5}{2}}\right)\right] \)
\( =\frac{1}{10}\left[\frac{\pi}{4}-\tan ^{-1} \frac{1}{\left(3^{\frac{5}{2}}\right)}\right] \)
\( \therefore I =\frac{1}{10}\left[\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right]\)