Area below the curve $y=\left[\sqrt{2+2\cos2x}\right]$ Find the area below the curve $y=[\sqrt{2+2\cos2x}]$ and above the $x$-axis , $x\in [-3\pi,6\pi]$, (where $[.]$ denotes the greatest integer function) .
My approach:
$$y=[\sqrt{2+2\cos2x}]$$
$$=[\sqrt{2+2(2\cos^2x-1}]$$
$$=[\sqrt{2+4\cos^2x-2}]$$
$$=[|2\cos x|]$$
After this, I do not know how to solve it and how to find the area. 
 A: $$0\le 2|\cos{x}|\le 2\implies [2|\cos{x}|]\in[0,1,2]$$
The function $[2|\cos{x}|]$ is discontinuous for those values of x that make it 2, but those are single points of removable discontinuity and they're not going to affect the integration properties of this function. So, the area lies between values of $0$ and $1$. Therefore, we need to find that value of x (just find one such value, the rest of the problem is going to be take care of by symmetry and repetition) which serves as the boundary point between values that are above 1 and values that are below 1:
$$
2\cos{x}=1\implies x=\frac{\pi}{3}
$$
Using symmetry and the fact that the area is just a bunch of rectangles that are the same, we have the following:
$$
\int_{-3\pi}^{6\pi}\sqrt{2+2\cos(2x)}\,dx=
\int_{-3\pi}^{6\pi}[|2\cos{x}|]\,dx=
18\int_{0}^{\frac{\pi}{3}}\,dx=6\pi\ sq.\ units.
$$
A: The values of $y$ are $0,\,1,\,2$. The last of these is achieved only on a measure-$0$ set irrelevant to integration. Thus $\int_{-3\pi}^{6\pi}ydx$ is the sum of the lengths of intervals on which $y\ge 1$ (i.e. $|\cos x|\ge\frac12$). With respect to $[-3\pi,\,6\pi]$, the complement of these intervals' union is the union of intervals over which $\cos x$ bijects from $\pm\frac12$ to $\mp\frac12$. There are nine such intervals, $\left(k\pi-\frac{8\pi}{3},\,k\pi-\frac{7\pi}{3}\right)$ for integer $k$ from $0$ to $8$ inclusive. As their total length is $9\times\frac{\pi}{3}=3\pi$, the sought integral is $9\pi-3\pi=6\pi$.
