JEE Main 2019 — Definite Integration Question with Solution

$\mathrm{I}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{dx}{\sin 2x\left({\tan }^{5}x+{\cot }^{5}x\right)}$ $\begin{array}{l} =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{5} x \cdot \sec ^{2} x}{2 \frac{\sin x}{\cos x}\left(\left(\tan ^{5} x\right)^{2}+1\right)} \\ =\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4} \tan ^{4} x \cdot \sec ^{2} x}{\left(\tan ^{5} x\right)^{2}+1} d x \end{array}$ Let ${\tan }^{4}x=t$ $5{\tan }^{4}x\cdot {\sec }^{2}xdx=dt$ When $x\to \frac{\pi }{4}$ then $t\to 1$ and $x\to \frac{\pi }{6}$ then $t\to {\left(\frac{1}{\sqrt{3}}\right)}^5$ $\begin{aligned} \therefore \quad & \mathrm{I}=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{d t}{t^{2}+1} \\ &=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right) \end{aligned}$