Mathematics - Definite Integration Question with Solution | TestHub

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