Question

If the value of ${\int }_0^{100\pi }\left(\right[{\cot }^{-1}x]+[{\tan }^{-1}x\left]\right)dx$ is $100\pi +p\cot p$, then the value of $p$ is

Note: $[.]$ denotes greatest integer function

Answer 1

Let $I={\int }_0^{100\pi }\left(\right[{\cot }^{-1}x]+[{\tan }^{-1}x\left]\right)dx$
$\Rightarrow I={\int }_0^{\cot 1}\left(1\right)dx+{\int }_{\cot 1}^{100\pi }\left(0\right)dx+{\int }_0^{\tan 1}\left(0\right)dx{\int }_{\tan 1}^{100\pi }\left(1\right)dx$
$\Rightarrow I=\cot 1+(100\pi -\tan 1)=100\pi +\frac{1-{\tan }^{2}1}{\tan 1}$

$\Rightarrow I=100\pi +2\frac{1-{\tan }^{2}1}{2\tan 1}=100\pi +2\frac{1}{\tan (1+1)}=100\pi +2\cot 2$

Comparing with $I=100\pi +p\cot p$, we get $p=2$

Hence, answer is $2$.