Question

Evaluate: $I={\int }_{-2}^1({\tan }^{-1}x+{\cot }^{-1}x)dx$

• A. $\frac{3\pi }{2}$
• B. $\frac{5\pi }{2}$
• C. $0$
• D. None of these

Answer 1

Answer: (A)

$\frac{3\pi }{2}$

$I$$=$${\int }_{-2}^1\left({\tan }^{-1}x+{\cot }^{-1}x\right)dx$

$={\int }_{-2}^1\frac{\pi }{2}dx$ (since $\text{ta}{\text{n}}^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$ $\forall x\in R$)

$=\frac{\pi }{2}{\left(x\right)}_{-2}^1$

$=\frac{\pi }{2}\left(1+2\right)$

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