Question:

Step 1: Determine the value of $[\cos x]$ over subintervals of $[0, \pi]$. The greatest integer function $[y]$ takes integer values. The range of $\cos x$ on $[0, \pi]$ is $[-1, 1]$. \begin{itemize} \item For $x \in [0, \pi/2)$: $\cos x \in (0, 1]$. At $x=0$, $\cos x = 1$, so $[\cos x] = 1$. For $x \in (0, \pi/2)$, $\cos x \in (0, 1)$, so $[\cos x] = 0$. The contribution of a single point to the integral is zero. Thus, for the purpose of integration over $[0, \pi/2)$, $[\cos x] = 0$. \item For $x \in [\pi/2, \pi]$: $\cos x \in [-1, 0]$. At $x=\pi/2$, $\cos x = 0$, so $[\cos x] = 0$. For $x \in (\pi/2, \pi)$, $\cos x \in (-1, 0)$, so $[\cos x] = -1$. At $x=\pi$, $\cos x = -1$, so $[\cos x] = -1$. Thus, for the purpose of integration over $(\pi/2, \pi]$, $[\cos x] = -1$. \end{itemize} Step 2: Evaluate the integral. The integral can be split based on these intervals: $$ \int_0^{\pi} [\cos x] dx = \int_0^{\pi/2} [\cos x] dx + \int_{\pi/2}^{\pi} [\cos x] dx $$ Substituting the values of $[\cos x]$ determined in Step 1: $$ = \int_0^{\pi/2} 0 \, dx + \int_{\pi/2}^{\pi} (-1) \, dx $$ $$ = 0 + (-1) \left[ x \right]_{\pi/2}^{\pi} $$ $$ = (-1) \left( \pi - \frac{\pi}{2} \right) $$ $$ = (-1) \left( \frac{\pi}{2} \right) $$ $$ = -\frac{\pi}{2} $$