The integral $\int_{\pi / 6}^{\pi / 4} \frac{\mathrm{d} x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$…
The integral $\int_{\pi / 6}^{\pi / 4} \frac{\mathrm{d} x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$ equals:
- $\frac{1}{20} \tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)$
- $\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$
- $\frac{\pi}{40}$
- $\frac{1}{5}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$
Solution
$\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$
$\begin{array}{l}
=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{5} x \cdot \sec ^{2} x}{2 \frac{\sin x}{\cos x}\left(\left(\tan ^{5} x\right)^{2}+1\right)} \\
=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4} \tan ^{4} x \cdot \sec ^{2} x}{\left(\tan ^{5} x\right)^{2}+1} d x
\end{array}$
Let $\tan ^{4} x=t$
$5 \tan ^{4} x \cdot \sec ^{2} x d x=d t$
When $x \rightarrow \frac{\pi}{4}$ then $t \rightarrow 1$
and $x \rightarrow \frac{\pi}{6}$ then $t \rightarrow\left(\frac{1}{\sqrt{3}}\right)^{5}$
$\begin{aligned}
\therefore \quad & \mathrm{I}=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{d t}{t^{2}+1} \\
&=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)
\end{aligned}$
Asked in: JEE Main 2019 (11 Jan Shift 2)
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