Question

The value of the definite integral $\underset{0}{\overset{\pi /2}{\int }}\left({\tan }^{-1}(\cot x)+{\cot }^{-1}(\tan x)\right)dx$ is equal to

• A. $\frac{{\pi }^2}{2}$
• B. $\frac{{\pi }^2}{4}$
• C. $\frac{{\pi }^2}{8}$
• D. ${\pi }^2$

Answer 1

Answer: (B)

Let $I=\underset{0}{\overset{\pi /2}{\int }}\left({\tan }^{-1}(\cot x)+{\cot }^{-1}(\tan x)\right)dx$
Using king
$I=\underset{0}{\overset{\pi /2}{\int }}\left({\tan }^{-1}(\tan x)+{\cot }^{-1}(\cot x)\right)dx\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(2x)dx={\left[x^2\right]}_0^{\pi /2}=\frac{{\pi }^2}{4}$