min{} in the integral I actually don't know what min{} in the question means. After knowing that I think I can solve it out.
$$ Q.\qquad \int_0^{2\pi}\min \left\{  |x-\pi|, \cos^{-1}(\cos x) \right\} \,dx $$
 A: $\min\{a,b,\ldots,c\}$ is the least element of the set $\{a,b,\ldots,c\}$. (A least element always exists because the set is finite).
A: Simplifying each function, you have that
  $|x-\pi|= \begin{cases}
\pi-x,  & \text{if $0\le x\le\pi$} \\
x-\pi, & \text{if $\pi\le x\le2\pi$}  \\
\end{cases}$
and $\cos^{-1}(\cos x)=\begin{cases}
x, & \text{if $0\le x\le\pi$} \\
2\pi-x, & \text{if $\pi\le x\le2\pi$.}  \\
\end{cases}$
Therefore $\min\big\{|x-\pi|,\cos^{-1}(\cos x)\big\}= \begin{cases}
x, & \text{if $0\le x\le \frac{\pi}{2}$} \\
\pi-x, & \text{if $\frac{\pi}{2}\le x\le \pi$} \\
x-\pi, & \text{if $\pi\le x\le\frac{3\pi}{2}$} \\
2\pi-x, & \text{if $\frac{3\pi}{2}\le x\le 2\pi.$} \\
\end{cases}$
(Notice that the integrals for each of these four intervals are the same,
by symmetry.)
A: Find the set $S \subseteq [0,2\pi]$ where $|x-\pi| < \cos^{-1}(\cos x)$.
Let $T = [0,2\pi] \setminus S$.
Then, the answer to your question is the sum $\int_S |x-\pi|dx + \int_T \cos^{-1}(\cos x) dx $ 
A: well $min()$ means functions inside the argument of it which has minimum value in given interval for example if we are given $min(x,x^2)$ now if we want to see it's value from 0 to infinity it this function will be $x^2$ in $[0,1]$ and $x$ thereafter since $x^2 < x $ in $(0,1)$.
