Evaluate $\int _0^{102} [\tan ^{-1} x].dx$ $$I=\int_{0}^{\tan 1} [\tan^{-1} x] + \int_{\tan 1}^{102} [\tan^{-1} x ]$$
Now the first integral is obviously zero. But I don’t know what the to do with the second one. Basically, I don’t know how to resolve floor functions in integrals. Can I get a hint?
One technique for approaching floor functions is looking at the graph of a function.  You noticed that between $0$ and $\tan 1$, the floor function is $0$.  On the other hand, the floor function between $\tan 1$ and $102$ is just $1$, due to the asymptotics of the tangent function (i.e. $1&lt; \pi/2 &lt; 2$).

Therefore,
$$\int_{\tan 1}^{102} [\tan^{-1} x ] \,dx = \int_{\tan 1}^{102} 1 \,dx = 102 - \tan 1.$$
If $[.]$ is GIF then, take $x= \tan t \implies dx= \sec^2t dt$
$$I=\int_{0}^{102} [\tan^{-1} x] dx= \int_{0}^{\tan^{-1} 102} [t] \sec^2 t dt$$
$$\implies I=\int_{0}^{1} [t] \sec^2 t dt+ \int_{1}^{\tan^{1} 102} [\pi/2- \epsilon]\sec^2 t dt= 0+\int_{1}^{\tan^{-1}{102}} 1 \sec^2 t dt=\tan t|_{1}^{\tan^{-1}102}$$
$$\implies I= \tan \tan^{-1} 102-\tan 1= 102-\tan 1.$$