Step 1: Find the derivative of the function. We have $f(x) = \cos^2 x + \sin x$. Differentiating with respect to $x$, we get: $f'(x) = 2 \cos x (-\sin x) + \cos x$ $f'(x) = -2 \sin x \cos x + \cos x$ $f'(x) = \cos x (1 - 2 \sin x)$.

Step 2: Find the critical points. Set $f'(x) = 0$. This gives $\cos x = 0$ or $1 - 2 \sin x = 0$. For $\cos x = 0$ in $x \in [0, \pi]$, we have $x = \frac{\pi}{2}$. For $1 - 2 \sin x = 0$, we have $\sin x = \frac{1}{2}$. In $x \in [0, \pi]$, this gives $x = \frac{\pi}{6}$. Thus, the critical points in the interval are $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$.

Step 3: Evaluate the function at critical points and endpoints. We evaluate $f(x)$ at $x = 0, \frac{\pi}{6}, \frac{\pi}{2}, \pi$.

1. At endpoint $x = 0$: $f(0) = \cos^2 0 + \sin 0 = 1 + 0 = 1$.
2. At critical point $x = \frac{\pi}{6}$: $f\left(\frac{\pi}{6}\right) = \cos^2 \frac{\pi}{6} + \sin \frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^2 + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$.
3. At critical point $x = \frac{\pi}{2}$: $f\left(\frac{\pi}{2}\right) = \cos^2 \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1$.
4. At endpoint $x = \pi$: $f(\pi) = \cos^2 \pi + \sin \pi = (-1)^2 + 0 = 1$.

Step 4: Compare the values. The values obtained are $1, \frac{5}{4}, 1, 1$. The maximum value is $\frac{5}{4}$ and the minimum value is $1$. Thus, the absolute maximum value is $\frac{5}{4}$ occurring at $x = \frac{\pi}{6}$, and the absolute minimum value is $1$ occurring at $x = 0, \frac{\pi}{2}, \pi$.