The value of the integral


The value of the integral $\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$

(where $[x]$ denotes the greatest integer less than or equal to x) is:


  1. (1) 0

  2. (2) $\sin 4$

  3. (3) 4

  4. (4) $4-\sin 4$

Correct Option: 1


$\operatorname{Let} f(x)=\frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}}$

So, $f(-x)=\frac{\sin ^{2}(-x)}{\left[\frac{-x}{\pi}\right]+\frac{1}{2}}$ $\because[-x]=-1-[x]$

$\Rightarrow f(-x)=\frac{\sin ^{2} x}{-1-\left[\frac{x}{\pi}\right]+\frac{1}{2}}=\frac{\sin ^{2} x}{-\frac{1}{2}-\left[\frac{x}{\pi}\right]}=-f(x)$

$\Rightarrow f(x)$ is odd function

Hence, $\int_{-2}^{2} f(x) d x=0$

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