Answer
$\sqrt 2\pi$
Work Step by Step
Given: $r=\sqrt {1+\cos 2 \theta}$
$r'=(\dfrac{1}{2}) (1+\cos 2 \theta)^{-1/2}(-2 \sin 2 \theta)$
The length is given as follows:
$L= \int_{-\pi/2}^{\pi/2} \sqrt{r^2+r'^2} d\theta=\int_{-\pi/2}^{\pi/2}\sqrt {2} d\theta$
Then, $L=(\sqrt 2)[\dfrac{\pi}{2}-(-\dfrac{\pi}{2})]_{-\pi/2}^{\pi/2} $
After solving, we get $L=\sqrt 2\pi$
