Question

$\underset{0}{\stackrel{5\pi }{\int }}\left[{\tan }^{-1}x\right]dx$ where [.] denotes the greatest integer function, is equal to

• A. $10\pi -\tan 1$
• B. $5\pi -\tan 1$
• C. $5-\tan 1$
• D. $10\pi +\tan 1$

Answer 1

The correct option is B $5\pi -\tan 1$

$I={\int }_0^{5\pi }\left[ta{n}^{-1}x\right]dx$

$\therefore \left[{\tan }^{-1}x\right]=0$ if $x\in (-tan,tan1)$

$\left[ta{n}^{-1}x\right]d=1$ if $x\in [tan1,\infty ]$

$I={\int }_0^{5\pi }\left[ta{n}^{-1}x\right]d={\int }_0^{tan1}\left[ta{n}^{-1}x\right]dx+{\int }_{tan1}^{5\pi }\left[ta{n}^{-1}x\right]dx$

$={\int }_0^{tan1}0dx+{\int }_{tan1}^{5\pi }dx$

$I=x{]}_{tan1}^{5\pi }$

$I=5\pi -tan1$

${\int }_0^{\pi }\left[ta{n}^{-1}x\right]dx=5\pi -tan1$