The area of the region enclosed by the curve $$f(x) = \max\{\sin x, \cos x\}$$, $$-\pi \leq x \leq \pi$$ and the $$x-$$axis is

We need to find the area enclosed by $$f(x) = \max\{\sin x, \cos x\}$$ for $$-\pi \leq x \leq \pi$$ and the $$x$$-axis.

Find where $$\sin x = \cos x$$

$$\sin x = \cos x$$ when $$\tan x = 1$$, i.e., $$x = \frac{\pi}{4}$$ and $$x = -\frac{3\pi}{4}$$ in $$[-\pi, \pi]$$.

Determine $$f(x)$$ on each interval

On $$[-\pi, -\frac{3\pi}{4}]$$: At $$x = -\pi$$, $$\sin(-\pi) = 0 > \cos(-\pi) = -1$$, so $$f(x) = \sin x$$.

On $$[-\frac{3\pi}{4}, \frac{\pi}{4}]$$: At $$x = 0$$, $$\cos 0 = 1 > \sin 0 = 0$$, so $$f(x) = \cos x$$.

On $$[\frac{\pi}{4}, \pi]$$: At $$x = \frac{\pi}{2}$$, $$\sin\frac{\pi}{2} = 1 > \cos\frac{\pi}{2} = 0$$, so $$f(x) = \sin x$$.

Compute the area (integral of $$|f(x)|$$)

Interval $$[-\pi, -\frac{3\pi}{4}]$$: Here $$f(x) = \sin x \leq 0$$, so $$|f(x)| = -\sin x$$.

$$\int_{-\pi}^{-3\pi/4} (-\sin x)\,dx = [\cos x]_{-\pi}^{-3\pi/4} = \cos\left(-\frac{3\pi}{4}\right) - \cos(-\pi) = -\frac{\sqrt{2}}{2} + 1 = 1 - \frac{\sqrt{2}}{2}$$

Interval $$[-\frac{3\pi}{4}, \frac{\pi}{4}]$$: Here $$f(x) = \cos x$$. Since $$\cos x < 0$$ on $$[-\frac{3\pi}{4}, -\frac{\pi}{2}]$$ and $$\cos x \geq 0$$ on $$[-\frac{\pi}{2}, \frac{\pi}{4}]$$, we split:

$$\int_{-3\pi/4}^{-\pi/2} (-\cos x)\,dx = [-\sin x]_{-3\pi/4}^{-\pi/2} = -\sin\left(-\frac{\pi}{2}\right) + \sin\left(-\frac{3\pi}{4}\right) = 1 - \frac{\sqrt{2}}{2}$$ $$\int_{-\pi/2}^{\pi/4} \cos x\,dx = [\sin x]_{-\pi/2}^{\pi/4} = \frac{\sqrt{2}}{2} - (-1) = \frac{\sqrt{2}}{2} + 1$$

Subtotal = $$\left(1 - \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{2}}{2} + 1\right) = 2$$

Interval $$[\frac{\pi}{4}, \pi]$$: Here $$f(x) = \sin x \geq 0$$.

$$\int_{\pi/4}^{\pi} \sin x\,dx = [-\cos x]_{\pi/4}^{\pi} = -\cos\pi + \cos\frac{\pi}{4} = 1 + \frac{\sqrt{2}}{2}$$

Total area

$$\text{Area} = \left(1 - \frac{\sqrt{2}}{2}\right) + 2 + \left(1 + \frac{\sqrt{2}}{2}\right) = 4$$

The correct answer is Option B: $$4$$.