The value of ∫limits_0^100 π Σlimits_r=1^10 tan r x d x is equal to

Let

$f(x)=\sum\limits_{r=1}^{10} \tan r x=\tan x+\tan 2 x+...+\tan 10 x$

Clearly, $f(x)$ is a periodic function with period $\pi$.

∴  $I=\int\limits_0^{100 \pi}\left(\sum\limits_{r=1}^{10} \tan r x\right) d x$

$\Rightarrow I=100 \int\limits_0^\pi\left(\sum\limits_{r=1}^{10} \tan r x\right) d x$

$\Rightarrow I=100\left[\sum\limits_{r=1}^{10} \int\limits_0^\pi \tan r x d x\right]$

∵  $\tan r(\pi-x)=-\tan r x$ for $r=1,2, ..., 10$

∴  $\int\limits_0^\pi \tan r x d x=0$

Hence, $I=100 \times \sum\limits_{r=1}^{10} 0=0$