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