If $$[x]$$ denotes the greatest integer less than or equal to $$x$$, then the value of the integral $$\int_{-\pi/2}^{\pi/2} [x] - \sin x] dx$$ is equal to:

We evaluate $$I = \int_{-\pi/2}^{\pi/2} \left[[x] - \sin x\right] dx$$, where $$[t]$$ denotes the greatest integer (floor) function.

We split the interval into four parts based on where $$[x]$$ is constant.

On $$\left(-\frac{\pi}{2}, -1\right)$$: $$[x] = -2$$, and $$\sin x \in (-1, -0.841)$$, so $$[x] - \sin x = -2 - \sin x \in (-1.159, -1)$$. Thus $$\left[[x] - \sin x\right] = -2$$. The contribution is $$(-2)\left(-1 + \frac{\pi}{2}\right) = 2 - \pi$$.

On $$[-1, 0)$$: $$[x] = -1$$, and $$\sin x \in (-0.841, 0)$$, so $$[x] - \sin x \in (-1, -0.159)$$. Thus $$\left[[x] - \sin x\right] = -1$$. The contribution is $$(-1)(1) = -1$$.

On $$[0, 1)$$: $$[x] = 0$$, and $$\sin x \in [0, \sin 1) \subset [0, 0.841)$$, so $$[x] - \sin x = -\sin x \in (-0.841, 0]$$. For $$x \in (0,1)$$, $$\left[[x] - \sin x\right] = -1$$. The contribution is $$(-1)(1) = -1$$.

On $$[1, \frac{\pi}{2}]$$: $$[x] = 1$$, and $$\sin x \in (\sin 1, 1] \subset (0.841, 1]$$, so $$[x] - \sin x = 1 - \sin x \in [0, 0.159)$$. Thus $$\left[[x] - \sin x\right] = 0$$. The contribution is $$0$$.

Summing all contributions: $$I = (2 - \pi) + (-1) + (-1) + 0 = -\pi.$$