# The integral

Question:

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

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

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

3. (3) $\frac{\pi}{40}$

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

Correct Option: , 2

Solution:

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

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{5} x \cdot \sec ^{2} x}{2 \frac{\sin x}{\cos x}\left(\left(\tan ^{5} x\right)^{2}+1\right)}$

$=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{4} x \cdot \sec ^{2} x}{\left(\tan ^{5} x\right)^{2}+1} d x$

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 $t \rightarrow\left(\frac{1}{\sqrt{3}}\right)^{5}$

$\therefore \quad \mathrm{I}=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{d t}{t^{2}+1}$

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