Integration with floor

The function $f(x) = \lfloor 2\sin(x) \rfloor$ takes on the following values over [ $-\pi, \frac{\pi}{2}$ ]:

$-2 \Rightarrow x = -\pi;$ (single point only!)

$-1 \Rightarrow x \in (-\pi, -\frac{5\pi}{6}] \cup [-\frac{\pi}{6}, 0)$ ;

$-2 \Rightarrow x \in (-\frac{5\pi}{6}, -\frac{\pi}{6});$

$0 \Rightarrow x \in [0, \frac{\pi}{6});$

$1 \Rightarrow x \in [\frac{\pi}{6}, \frac{\pi}{2}).$

$2 \Rightarrow x = \frac{\pi}{2}.$ (single point only!)

The integration can now proceed according to:

$\int_{-\pi}^{-5\pi/6} -1 dx +\int_{-5\pi/6}^{-\pi/6} -2 dx +\int_{-\pi/6}^{0} -1 dx +\int_{0}^{\pi/6} 0 dx + \int_{\pi/6}^{\pi/2} 1 dx$ = $\boxed{-\frac{4\pi}{3}}.$