Let the area of the region enclosed by the curve $$y = \min\{\sin x, \cos x\}$$ and the $$x$$ axis between $$x = -\pi$$ to $$x = \pi$$ be $$A$$. Then $$A^2$$ is equal to ________

Find $$A^2$$ where $$A$$ is the area between $$y = \min\{\sin x, \cos x\}$$ and the x-axis from $$-\pi$$ to $$\pi$$.

$$\sin x = \cos x$$ at $$x = -3\pi/4$$ and $$x = \pi/4$$. The function is:

$$[-\pi, -3\pi/4]$$: $$y = \cos x \leq 0$$. $$[-3\pi/4, \pi/4]$$: $$y = \sin x$$. $$[\pi/4, \pi]$$: $$y = \cos x$$.

Computing area (taking absolute value where function is negative):

$$[-\pi, -3\pi/4]$$: $$\int(-\cos x)dx = \frac{\sqrt{2}}{2}$$

$$[-3\pi/4, 0]$$: $$\int(-\sin x)dx = 1+\frac{\sqrt{2}}{2}$$

$$[0, \pi/4]$$: $$\int \sin x\,dx = 1-\frac{\sqrt{2}}{2}$$

$$[\pi/4, \pi/2]$$: $$\int \cos x\,dx = 1-\frac{\sqrt{2}}{2}$$

$$[\pi/2, \pi]$$: $$\int(-\cos x)dx = 1$$

Total: $$A = \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 = 4$$.

$$A^2 = \boxed{16}$$.