Practicing Success
The value of $\int\limits_0^{100 \pi} \sum\limits_{r=1}^{10} \tan r x d x$ is equal to |
$100 \pi$ $-100 \pi$ 1 none of these |
none of these |
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, $I=100 \times \sum\limits_{r=1}^{10} 0=0$ |