Solving Integral involving greatest integer function: $\int_{0}^{100}\left[\tan^{-1}x\right] dx$ and $\int_{0}^{2\pi}\left[\cot^{-1}x\right]dx $ 
Question Solve (a) $\int_{0}^{100}\left[\tan^{-1}x\right] dx$
$\left(b\right)$ $\int_{0}^{2\pi}\left[\cot^{-1}x\right]dx $

I have found a Helpful post
It suggest there are many other formulas like this that may help.Can anyone please tell me what are they or where i can find them ?
Book's Answer$\left(a\right)$$\int_{0}^{\tan1}\left[\tan^{-1}x\right]dx$+$\int_{\tan1}^{100}$$\left[\tan^{-1}x\right]$dx=$\int_{0}^{tan1}0.dx$+$\int_{\tan1}^{100}1dx$
But I cannot understand it what is happenning. Please Please Please
Suggest me study material or book. So I can do them myself
 A: The floor function, which is denoted by $\lfloor x \rfloor$, gives the greatest integer for the real number $x$. 
For the inverse tangent function on the interval $[0,100]$ its value lies between zero and its asymptotic value of $\pi/2$ which it approaches from below (this information comes from knowing what the graph of the inverse tangnet function looks like and is a very important skill to have). 
Knowing all this we now need to establish for what values of $x$ is $\tan^{-1} x$ is less than one and for what values of $x$ is $\tan^{-1} x$ greater than one for $x \in [0,100]$.
As the inverse tangent function monotonically increases from zero when $x = 0$ to a value between one and two when $x = 100$ (as $\tan^{-1} (100) < 2$), finding when $\tan^{-1} (x) = 1$ we see this occurs when $x = \tan 1$.
So for our integral we have
\begin{align*}
\int^{100}_0 \lfloor \tan^{-1} x \rfloor \, dx &= \int^{\tan 1}_0 \lfloor \tan^{-1} x \rfloor \, dx + \int^{100}_{\tan 1} \lfloor \tan^{-1} x \rfloor \, dx\\
&= \int^{\tan 1}_0 0 \, dx + \int^{100}_{\tan 1} 1 \, dx\\
&= 100 - \tan 1.
\end{align*}  
Your integral for the inverse cotangent function can be now be done in a similar fashion. 
