Answer
$1$
Work Step by Step
Given
\begin{aligned} \int_{0}^{\pi} \frac{1}{2}(\cos x+|\cos x|) d x\end{aligned}
Since
\begin{aligned} |\cos(x)|&= \begin{cases}\cos(x)&0\leq x\lt \pi /2\\
-\cos(x)&\pi/2\leq x\leq \pi
\end{cases}\end{aligned}
Then
\begin{aligned} \int_{0}^{\pi} \frac{1}{2}(\cos x+|\cos x|) d x&=
\int _0^{\frac{\pi }{2}}\frac{1}{2}\left(\cos \left(x\right)+\cos \left(x\right)\right)dx+\int _{\frac{\pi }{2}}^{\pi }\frac{1}{2}\left(\cos \left(x\right)-\cos \left(x\right)\right)dx\\
&=\int _0^{\frac{\pi }{2}}\left(\cos \left(x\right)\right)dx\\
&= \sin(x)\bigg|_0^{\pi/2}\\
&= 1
\end{aligned}
