Eye of the Wave

It is the x-coordinate of the intersection between the red and green lines is the solution to $\sin(x)=1-\sin(x)$ , $0≤x≤\pi$ , which is $x=\frac{\pi}{6}$ .

Using the same methods, we can find that the intersection between the blue and the red lines has an x-coordinate of $\frac{\pi}{4}$ , that between the purple and green lines $\frac{\pi}{4}$ , and that between the blue and purple lines $\frac{\pi}{3}$ .

As the shaded region is symmetric, it is double the area enclosed from $x=\frac{\pi}{6}$ to $x=\frac{\pi}{4}$ , which can be calculated by the integral of the red line minus the green line.

The shaded area is therefore

$2\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\sin(x)-(1-\sin(x)))dx$

$=2\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (2\sin(x)-1)dx$

$=2[-2\cos(x)-x]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

$=2(-2\frac{\sqrt{3}}{2}-\frac{\pi}{6}+2\frac{\sqrt{2}}{2}+\frac{\pi}{4})$

$=2(\sqrt{3}-\sqrt{2})-\frac{\pi}{6}$

So $A=2,B=3,C=2,D=6$ , giving the answer of $\boxed{13}$ .