Evaluate the following integrals:
Question:

The integral $\int_{\pi / 6}^{\pi / 4} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$ equals :-

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

2. $\frac{1}{5}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$

3. $\frac{\pi}{10}$

4. $\frac{1}{20} \tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)$

Correct Option: 1

Solution:

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

$I=\frac{1}{2} \int_{x / 6}^{\pi / 4} \frac{\tan ^{4} x \sec ^{2} x d x}{\left(1+\tan ^{10} x\right)}$ Put $\tan ^{5} x=t$

$I=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1} \frac{1}{9 \sqrt{3}}\right)$