The value of $\int\limits_0^{32 \pi / 3} \sqrt{1+\cos 2 x} d x$, is

$\frac{44-\sqrt{3}}{\sqrt{2}}$

We have,

$\int\limits_0^{32} \sqrt{1+\cos 2 x} d x$

$= \sqrt{2} \int\limits_0^{10 \pi}|\cos x| d x+\sqrt{2} \int\limits_{10 \pi}^{32 \pi / 3}|\cos x| d x$

$=10 \sqrt{2} \int\limits_0^\pi|\cos x| d x+\sqrt{2} \int\limits_0^{2 \pi / 3}|\cos x| d x$

$=10 \sqrt{2}\left[\int\limits_0^{\pi / 2} \cos x d x-\int\limits_{\pi / 2}^\pi \cos x d x\right] +\sqrt{2}\left[\int\limits_0^{\pi / 2} \cos x d x+\int\limits_{\pi / 2}^{2 \pi / 3}-\cos x d x\right]$

$=10 \sqrt{2}[1+1]+\sqrt{2}\left[1-\frac{\sqrt{3}}{2}+1\right]$

$=20 \sqrt{2}+\sqrt{2}\left(2-\frac{\sqrt{3}}{2}\right)=22 \sqrt{2}-\sqrt{\frac{3}{2}}=\frac{44-\sqrt{3}}{\sqrt{2}}$