The value of the integral $\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$
(where $[\mathrm{x}]$ denotes the greatest integer less than ${ }^{20} \mathrm{Cr}$ or equal to $\mathrm{x}$ ) is :
Correct Option: , 4
$I=\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$
$I=\int_{0}^{2}\left(\frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}}+\frac{\sin ^{2}(-x)}{\left[-\frac{x}{\pi}\right]+\frac{1}{2}}\right) d x$
$\left(\left[\frac{\mathrm{x}}{\pi}\right]+\left[-\frac{\mathrm{x}}{\pi}\right]=-1\right.$ as $\left.\quad \mathrm{x} \neq \mathrm{n} \pi\right)$
$I=\int_{0}^{2}\left(\frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}}+\frac{\sin ^{2} x}{-1-\left[\frac{x}{\pi}\right]+\frac{1}{2}}\right) d x=0$
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