Sketch the region enclosed by the given curves. $y = 4 \cos 6x$, $y = 4 − 4 \cos 6x$, $0 ≤ x ≤ π/6$. Find it's area. 
For this one, I believe that there are 2 area: 1 for the left and 1 for the right of the intersection point. I believe that I would had to add the area of those two together to get the final answer. However, I am not sure what I did Wrong.

For the 1st area (the one on the left), I use:
$b = \pi/8$
$a = 0$
$$\int [4\cos(6x) - (4-4\cos(6x))] dx$$

For the 2nd area (the one on the right), I use:
$b = \pi/6$
$a = \pi/8$
$$\int[ 4-4\cos(6x) - (4\cos(6x))] dx$$
 A: Hint: The two functions intersect when $4-4\cos(6x)=4\cos(6x)$, i.e., $8\cos(6x)=4$, or equivalently $\cos(6x)=1/2$. When $x\in [0,\pi/6]$, $x=\pi/18$ satisfies the intersection point.
A: Well, we can write the area as:
$$\mathcal{A}_\text{T}=\mathcal{A}_1+\mathcal{A}_2+\mathcal{A}_3\tag1$$
Where:
$$\mathcal{A}_1:=\int_0^\frac{\pi}{18}4\cos\left(6x\right)\space\text{d}x-\int_0^\frac{\pi}{18}\left(4-4\cos\left(6x\right)\right)\space\text{d}x\tag2$$
$$\mathcal{A}_2:=\int_\frac{\pi}{18}^\frac{\pi}{12}\left(4-4\cos\left(6x\right)\right)\space\text{d}x-\int_\frac{\pi}{18}^\frac{\pi}{12}4\cos\left(6x\right)\space\text{d}x\tag3$$
$$\mathcal{A}_3:=\int_\frac{\pi}{12}^\frac{\pi}{6}\left(4-4\cos\left(6x\right)\right)\space\text{d}x+\left|\int_\frac{\pi}{12}^\frac{\pi}{6}4\cos\left(6x\right)\space\text{d}x\right|\tag4$$
A: Hint :
The graph is symmetric about $y=2$ and intersection point is on $x=\pi/18$. So evaluate the following integrals :
$$2\times \left[ \int_0^{\pi/18} (4\cos 6x-2) \, dx + \int_{\pi/18}^{\pi/6} (2-4\cos 6x) \, dx \right]$$
