1. We need to evaluate $I=\int_{-2}^{2}\frac{\sin^2 x}{\left[\frac{x}{\pi}\right]+\frac12}\,dx.$ Here $[\cdot]$ is the greatest integer function.

2. First, determine the value of $\left[\frac{x}{\pi}\right]$ on the interval $[-2,2]$.

Since $\pi\approx 3.14$, we have $-\pi < -2 \le x \le 2 < \pi.$ So:

• for $x\in[-2,0)$, we have $-1<\frac{x}{\pi}<0$, hence $\left[\frac{x}{\pi}\right]=-1;$
• at $x=0$ and for $x\in[0,2]$, we have $0\le \frac{x}{\pi}<1$, hence $\left[\frac{x}{\pi}\right]=0.$

Thus the denominator becomes:

• on $[-2,0)$: $\left[\frac{x}{\pi}\right]+\frac12=-1+\frac12=-\frac12;$
• on $[0,2]$: $\left[\frac{x}{\pi}\right]+\frac12=0+\frac12=\frac12.$

3. Split the integral at $x=0$: $I=\int_{-2}^{0}\frac{\sin^2 x}{-1/2}\,dx+\int_{0}^{2}\frac{\sin^2 x}{1/2}\,dx.$ So, $I=-2\int_{-2}^{0}\sin^2 x\,dx+2\int_{0}^{2}\sin^2 x\,dx.$

4. Now use the fact that $\sin^2 x$ is an even function: $\sin^2(-x)=\sin^2 x.$ Therefore, $\int_{-2}^{0}\sin^2 x\,dx=\int_{0}^{2}\sin^2 x\,dx.$ Let this common value be $A$. Then $I=-2A+2A=0.$

5. Hence, $\boxed{I=0}.$

6. Option check:

• A: $0$ ✅
• B: $4$ ❌
• C: $4-\sin 4$ ❌
• D: $\sin 4$ ❌

Therefore the correct option is A.