Answer
The minimum value of $f(x)$ is:
$f(\pi)=\frac{\pi}{2}. (-1)= \frac{-\pi}{2}$
Work Step by Step
$$
f(x) =\int_{0}^{\pi} \cos t \cos (x-t) d t\
$$
we need the following formula:
$$
\cos x \cos y=\frac{1}{2} \left[ \cos (x+y) +\cos (x-y)\right]
$$
Thus, we have
$$
\begin{aligned} f(x) &=\int_{0}^{\pi} \cos t \cos (x-t) d t\\
& =\frac{1}{2} \int_{0}^{\pi}[\cos (t+x-t)+\cos (t-x+t)] d t \\
&=\frac{1}{2} \int_{0}^{\pi}[\cos x+\cos (2 t-x)] d t \\
&=\frac{1}{2}\left[t \cos x+\frac{1}{2} \sin (2 t-x)\right]_{0}^{\pi} \\
&=\frac{\pi}{2} \cos x+\frac{1}{4} \sin (2 \pi-x)-\frac{1}{4} \sin (-x) \\ &=\frac{\pi}{2} \cos x+\frac{1}{4} \sin (-x)-\frac{1}{4} \sin (-x) \\
&=\frac{\pi}{2} \cos x \end{aligned}
$$
since the domain of the function $ D_ {\cos x} \in[-1,1]$, the the minimum value of the function $\cos x$ is equal to -1. So, the minimum value of $f(x)$ is:
$f(\pi)=\frac{\pi}{2}. (-1)= \frac{-\pi}{2}$
