Question

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

Answer 1

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan^2 x + \cot^2 x + 2\tan x\cot x \right)dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \sec^2 x - 1 + {cosec}^2 x - 1 + 2 \right)dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \sec^2 xdx + \int_\frac{\pi}{6}^\frac{\pi}{3} {cosec}^2 xdx\]

\[= \left.\tan x\right|_\frac{\pi}{6}^\frac{\pi}{3} + \left.\left( - \cot x \right)\right|_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \left( \tan\frac{\pi}{3} - \tan\frac{\pi}{6} \right) - \left( \cot\frac{\pi}{3} - \cot\frac{\pi}{6} \right)\]
\[ = \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) - \left( \frac{1}{\sqrt{3}} - \sqrt{3} \right)\]
\[ = 2\sqrt{3} - \frac{2}{\sqrt{3}}\]
\[ = \frac{4}{\sqrt{3}}\]