The integral $\int_{\pi / 6}^{\pi / 4} \frac{\mathrm{d} x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$ equals :
Correct Option: , 2
$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{5} x \cdot \sec ^{2} x}{2 \frac{\sin x}{\cos x}\left(\left(\tan ^{5} x\right)^{2}+1\right)}$
$=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{4} x \cdot \sec ^{2} x}{\left(\tan ^{5} x\right)^{2}+1} d x$
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 $t \rightarrow\left(\frac{1}{\sqrt{3}}\right)^{5}$
$\therefore \quad \mathrm{I}=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{d t}{t^{2}+1}$
$=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$
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