$\text{ The value of }{\int }_0^{100\pi } \left(\sum _{r=1}^{10} \tan rx\right)\mathrm{d}x\text{ is equalto }$

$\begin{array}{l}\text{ Let }I={\int }_0^{100\pi } \sum _{r=1}^{10} (\tan rx)\mathrm{d}x \\ ={\int }_0^{100\pi } [\tan x+\tan 2x+\tan 3x+\dots \dots +\tan 10x]\mathrm{d}x\\ I=100{\int }_0^{\pi } [\tan x+\tan 2x+\dots .+\tan (10x\left)\right]\mathrm{d}x \left(\sin ce {\int }_0^{nT}f\left(x\right)dx = n{\int }_0^{T}f\left(x\right)dx if f\left(x\right) period is T \right)\\ \Rightarrow I=-100{\int }_0^{\pi } (\tan x+\tan 2x+\dots \dots ..+\tan 10x)\mathrm{d}x \left(\sin ce {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right)\end{array}$

$therefore I +I =0 \Rightarrow I =0$