If [ ] denotes the greatest integer function, then the integral $$\int_0^{\pi} [\cos x] dx$$ is equal to:

The integral to evaluate is $$\int_0^{\pi} [\cos x] dx$$, where $$[ \cdot ]$$ denotes the greatest integer function. This function returns the greatest integer less than or equal to its argument. The cosine function, $$\cos x$$, decreases from 1 to -1 as $$x$$ goes from 0 to $$\pi$$. The greatest integer function $$[ \cos x ]$$ will change values at points where $$\cos x$$ is an integer, specifically at $$x = 0$$, $$x = \pi/2$$, and $$x = \pi$$, because $$\cos x$$ takes integer values 1, 0, and -1 at these points, respectively. To compute the integral, split it at the discontinuity points, specifically at $$x = \pi/2$$: $$$ \int_0^{\pi} [\cos x] dx = \int_0^{\pi/2} [\cos x] dx + \int_{\pi/2}^{\pi} [\cos x] dx $$$ Now, evaluate each integral separately. First, consider $$\int_0^{\pi/2} [\cos x] dx$$. - At $$x = 0$$, $$\cos 0 = 1$$, so $$[ \cos 0 ] = [1] = 1$$. - For $$x \in (0, \pi/2]$$, $$\cos x$$ is in the interval $$(0, 1]$$. Since $$\cos x \gt 0$$ and $$\cos x \lt 1$$ for $$x \gt 0$$, the greatest integer less than or equal to $$\cos x$$ is 0. At $$x = \pi/2$$, $$\cos(\pi/2) = 0$$, so $$[0] = 0$$. Thus, in the interval $$[0, \pi/2]$$, $$[ \cos x ] = 1$$ only at the single point $$x = 0$$, and $$[ \cos x ] = 0$$ for all $$x \in (0, \pi/2]$$. In the Riemann integral, the value at a single point does not affect the result because it has measure zero. Therefore, the function $$[ \cos x ]$$ is effectively 0 almost everywhere in $$[0, \pi/2]$$, and the integral is: $$$ \int_0^{\pi/2} [\cos x] dx = \int_0^{\pi/2} 0 dx = 0 $$$ To be rigorous, consider the limit as $$\epsilon \to 0^+$$: $$$ \int_0^{\pi/2} [\cos x] dx = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{\pi/2} [\cos x] dx $$$ For $$\epsilon \gt 0$$ and small, $$x \in [\epsilon, \pi/2]$$, $$\cos x \in (0, 1)$$, so $$[ \cos x ] = 0$$. Thus: $$$ \int_{\epsilon}^{\pi/2} 0 dx = 0 $$$ As $$\epsilon \to 0^+$$, the integral remains 0. Next, consider $$\int_{\pi/2}^{\pi} [\cos x] dx$$. - At $$x = \pi/2$$, $$\cos(\pi/2) = 0$$, so $$[0] = 0$$. - For $$x \in (\pi/2, \pi]$$, $$\cos x$$ is in the interval $$(-1, 0)$$. For example, at $$x = \pi$$, $$\cos \pi = -1$$, so $$[-1] = -1$$. For any $$y \in (-1, 0)$$, the greatest integer less than or equal to $$y$$ is $$-1$$ (e.g., $$[-0.5] = -1$$). Thus, in the interval $$[\pi/2, \pi]$$, $$[ \cos x ] = 0$$ only at the single point $$x = \pi/2$$, and $$[ \cos x ] = -1$$ for all $$x \in (\pi/2, \pi]$$. Again, the value at a single point does not affect the integral, so $$[ \cos x ]$$ is effectively $$-1$$ almost everywhere in $$[\pi/2, \pi]$$. Therefore: $$$ \int_{\pi/2}^{\pi} [\cos x] dx = \int_{\pi/2}^{\pi} (-1) dx = -1 \cdot \left( \pi - \frac{\pi}{2} \right) = -1 \cdot \frac{\pi}{2} = -\frac{\pi}{2} $$$ To be rigorous, consider the limit as $$\delta \to 0^+$$: $$$ \int_{\pi/2}^{\pi} [\cos x] dx = \lim_{\delta \to 0^+} \left( \int_{\pi/2 + \delta}^{\pi} [\cos x] dx + \int_{\pi/2}^{\pi/2 + \delta} [\cos x] dx \right) $$$ For $$x \in [\pi/2 + \delta, \pi]$$, $$\cos x \in [-1, -\sin \delta) \subset (-1, 0)$$, so $$[ \cos x ] = -1$$. Thus: $$$ \int_{\pi/2 + \delta}^{\pi} (-1) dx = -1 \cdot \left( \pi - \left( \frac{\pi}{2} + \delta \right) \right) = -\left( \frac{\pi}{2} - \delta \right) $$$ For $$x \in [\pi/2, \pi/2 + \delta]$$, $$[ \cos x ]$$ is bounded between $$-1$$ and $$0$$, so the integral is bounded in absolute value by $$\delta \cdot 1 = \delta$$. As $$\delta \to 0^+$$, this part approaches 0. Thus: $$$ \lim_{\delta \to 0^+} \left( -\left( \frac{\pi}{2} - \delta \right) + \text{bounded term} \right) = -\frac{\pi}{2} $$$ Now, sum the two integrals: $$$ \int_0^{\pi} [\cos x] dx = \int_0^{\pi/2} [\cos x] dx + \int_{\pi/2}^{\pi} [\cos x] dx = 0 + \left( -\frac{\pi}{2} \right) = -\frac{\pi}{2} $$$ Therefore, the integral evaluates to $$-\pi/2$$. Hence, the correct answer is Option D.

So, the answer is $$-\frac{\pi}{2}$$.