## University Calculus: Early Transcendentals (3rd Edition)

$\sqrt 2\pi$
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 $L=\sqrt 2\pi$