Answer
$2 \pi$
Work Step by Step
Given
$$
r=2 \cos \theta, \quad 0 \leqslant \theta \leqslant \pi
$$
The length is given by
\begin{align*}
L&=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{ d \theta}\right)^{2}} d \theta\\
&=\int_{0}^{\pi} \sqrt{(2 \cos \theta)^{2}+(-2 \sin \theta)^{2}} d \theta\\
&= \int_{0}^{\pi} \sqrt{4\left(\cos ^{2} \theta+\sin ^{2} \theta\right)} d \theta\\
&=\int_{0}^{\pi} \sqrt{4} d \theta=[2 \theta]_{0}^{\pi}\\
&=2 \pi
\end{align*}
