Question

$\underset{\mathrm{\pi }/3}{\overset{\mathrm{\pi }/4}{\int }}{\left(\tan x+\cot x\right)}^{2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{4}}{\left(\tan x+\cot x\right)}^{2}dx.\mathrm{Then}, \\ I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{4}}\left({\tan }^{2}x+{\cot }^{2}x+2\tan x\cot x\right)dx \\ \Rightarrow I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{4}}\left({\tan }^{2}x+{\cot }^{2}x+2\right)dx \\ \Rightarrow I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{4}}\left({\sec }^{2}x-1+{\mathrm{cosec}}^{2}x-1+2\right)dx \\ \Rightarrow I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{4}}\left({\sec }^{2}x+{\mathrm{cosec}}^{2}x\right)dx \\ \Rightarrow I={\left[\tan x-\cot x\right]}_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{4}} \\ \Rightarrow I=\left(1-1\right)-\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) \\ \Rightarrow I=\frac{-2}{\sqrt{3}} \end{gathered}$$