Question

The value of ${\int }_{-2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx$ is equal to

• A. $\frac{7\pi }{2}$
• B. $\frac{7{\pi }^2}{2}$
• C. $\frac{3\pi }{2}$
• D. ${\pi }^2$

Answer 1

The correct option is B $\frac{7{\pi }^2}{2}$

${\int }_{-2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx={\int }_{-2\pi }^{5\pi }{\cot }^{-1}\left(\cot (\frac{\pi }{2}-x)\right)dx$
$=7{\int }_0^{\pi }{\cot }^{-1}\left(\cot (\frac{\pi }{2}-x)\right)dx$
$=7\left\{{\int }_0^{\frac{\pi }{2}}(\frac{\pi }{2}-x)dx+{\int }_{\frac{\pi }{2}}^{\pi }(\frac{3\pi }{2}-x)dx+\right\}$
$=7\{{(\frac{x\pi }{2}-\frac{x^2}{2})}_0^{\frac{\pi }{2}}+{(\frac{3x\pi }{2}-\frac{x^2}{2})}_{\frac{\pi }{2}}^{\pi }\}=\frac{7{\pi }^2}{2}$