Question

${\int }_{\pi /3}^{\pi /4}(\tan x+\cot x{)}^{2}dx$.

Answer 1

$={\int }_{\pi /3}^{\pi /21}(\tan x+\cot x{)}^2$

$={\int }_{\pi /3}^{\pi /4}{\tan }^{2}x+{\cot }^{2}x+2$

$={\int }_{\pi /3}^{\pi /4}{\sec }^{2}x-1+{\csc }^2\left(x\right)-1+2$

$={\int }_{\pi /4}^{\pi /3}{\sec }^{2}x+{\csc }^2\left(x\right)$

$[\tan x-\cot x{]}_{\pi /3}^{\pi /4}$

$\Rightarrow [1-1]-[\sqrt{3}-\frac{1}{\sqrt{3}}]$

$=-\left(\frac{3-1}{\sqrt{3}}\right)=\frac{-2}{\sqrt{3}}$