Question

$\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}{(\tan x+\cot x)}^{2}dx$

Answer 1

$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}{(\tan x+\cos x)}^{2}dx$

$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}({\tan }^{2}x+{\cot }^{2}x+2\tan x\cot x)dx$

$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}({\sec }^{2}x-1+{\csc }^{2}x-1+2)dx$

$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}({\sec }^{2}x+{\csc }^{2}x)dx$

$={[\tan x-\cot x]}_{\frac{\pi }{6}}^{\frac{\pi }{4}}$

$=(\tan \frac{\pi }{4}-\cot \frac{\pi }{4})-(\tan \frac{\pi }{6}-\cot \frac{\pi }{6})$

$=(1-1)-(\frac{1}{\sqrt{3}}-\sqrt{3})$

$=-\frac{1-3}{\sqrt{3}}$

$=\frac{2}{\sqrt{3}}$