If for a real number $y,[y]$ is the greatest integer less than or equal to $y$, then the value of the integral $\int\limits_{\pi / 2}^{3 \pi / 2}[2 \sin x] d x$, is

$-\pi / 2$

We have,

$\int\limits_{\pi / 2}^{3 \pi / 2}[2 \sin x] d x=\int\limits_{\pi / 2}^{5 \pi / 6}[2 \sin x] d x+\int\limits_{5 \pi / 6}^\pi[2 \sin x] d x + \int\limits_\pi^{7 \pi / 6}[2 \sin x] d x+\int\limits_{7 \pi / 6}^{3 \pi / 2}[2 \sin x] d x$

$\Rightarrow \int\limits_{\pi / 2}^{3 \pi / 2}[2 \sin x] d x=\int\limits_{\pi / 2}^{5 \pi / 6} 1 d x+\int\limits_{5 \pi / 6}^\pi 0 d x+\int\limits_\pi^{7 \pi / 6}-1 d x+\int\limits_{7 \pi / 6}^{3 \pi / 2}-2 d x$

$\Rightarrow \int\limits_{\pi / 2}^{3 \pi / 2}[2 \sin x] d x=\left(\frac{5 \pi}{6}-\frac{\pi}{2}\right)-\left(\frac{7 \pi}{6}-\pi\right)-2\left(\frac{3 \pi}{2}-\frac{\pi}{6}\right)=-\frac{\pi}{2}$