The value of the integral $\underset{-2}{\overset{2}{\int }}\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+dx}$ (where$\left[x\right]$ denotes the greatest integer less than or equal to (x) is 

$$\begin{gathered} \mathrm{I}={\int }_{-2}^2 \frac{{\sin }^2 \mathrm{x}}{\left[{\displaystyle \frac{\mathrm{x}}{\mathrm{\pi }}}\right]+{\displaystyle \frac{1}{2}}} \\ ={\int }_{-2}^2 \frac{{\sin }^2 (-\mathrm{x})}{\left[{\displaystyle \frac{-\mathrm{x}}{\mathrm{\pi }}}\right]+{\displaystyle \frac{1}{2}}} \\ 2\mathrm{I}={\int }_{-2}^2 \frac{{\sin }^2 \mathrm{x}\left(\left[{\displaystyle \frac{\mathrm{x}}{\mathrm{\pi }}}\right]+{\displaystyle \frac{1}{2}}+\left[{\displaystyle \frac{-\mathrm{x}}{\mathrm{\pi }}}\right]{\displaystyle +\frac{1}{2}}\right)}{\left(\left[{\displaystyle \frac{\mathrm{x}}{\mathrm{\pi }}}\right]+{\displaystyle \frac{1}{2}}\right) \left(\left[{\displaystyle \frac{-\mathrm{x}}{\mathrm{\pi }}}\right]+\frac{1}{2}\right)} \\ 2\mathrm{I}={\int }_{-2}^2 \frac{{\sin }^2\mathrm{x} (1-1)}{\left(\left[{\displaystyle \frac{\mathrm{x}}{\mathrm{\pi }}}\right]+{\displaystyle \frac{1}{2}}\right) \left(\left[{\displaystyle \frac{\mathrm{x}}{\mathrm{\pi }}}\right]-{\displaystyle \frac{1}{2}}\right)} \\ 2\mathrm{I}=0 \\ \mathrm{I}=0 \\ (\because [\mathrm{x}]+[-\mathrm{x}]=0 \mathrm{if} \mathrm{x}\in \mathrm{Z} \\ = -1 \mathrm{if} \mathrm{x}\notin \mathrm{Z}) \end{gathered}$$