Question

Evaluate ${\int }_{\frac{\pi }{2}}^{2\pi }\left[co{t}^{-1}x\right]dx$, where $[\cdot ]$ denotes the greatest integer function.

Answer 1

Given : ${\int }_{\frac{\pi }{2}}^{2\pi }\left[{\cot }^{-1}x\right]dx$

$I={\int }_{\frac{\pi }{2}}^{2\pi }\left[{\cot }^{-1}x\right]dx$

Adding and subtracting ${\int }_0^{\frac{\pi }{2}}\left[{\cot }^{-1}x\right]dx$

$I={\int }_0^{\frac{\pi }{2}}\left[{\cot }^{-1}x\right]dx+{\int }_{\frac{\pi }{2}}^{2\pi }\left[{\cot }^{-1}x\right]dx-{\int }_0^{\frac{\pi }{2}}\left[{\cot }^{-1}x\right]dxI={\int }_0^{2\pi }\left[{\cot }^{-1}x\right]dx-{\int }_0^{\frac{\pi }{2}}\left[{\cot }^{-1}x\right]dx$

We know $\left[{\cot }^{-1}x\right]=\{10<x<\cot 1(0.642\left)\right\}$

$\left[{\cot }^{-1}x\right]=\{0x\ge \cot 1\}$

$I={\int }_0^{\cot 1}\left[{\cot }^{-1}x\right]dx+{\int }_{\cot 1}^{2\pi }\left[{\cot }^{-1}x\right]dx-{\int }_0^{\cot 1}\left[{\cot }^{-1}x\right]dx-{\int }_{\cot 1}^{\frac{\pi }{2}}\left[{\cot }^{-1}x\right]dxI=0+0-0-0I=0$

Hence the correct answer is $0$