Which of the following statement is correct for the function $$g(\alpha)$$ for $$\alpha \in R$$ such that $$g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^\alpha x}{\cos^\alpha x + \sin^\alpha x} dx$$:

We need to analyze $$g(\alpha) = \int_{\pi/6}^{\pi/3} \frac{\sin^\alpha x}{\cos^\alpha x + \sin^\alpha x}\, dx$$.

Using the substitution $$x = \frac{\pi}{2} - t$$, so $$dx = -dt$$. When $$x = \pi/6$$, $$t = \pi/3$$ and when $$x = \pi/3$$, $$t = \pi/6$$.

$$g(\alpha) = \int_{\pi/3}^{\pi/6} \frac{\sin^\alpha(\pi/2 - t)}{\cos^\alpha(\pi/2 - t) + \sin^\alpha(\pi/2 - t)} (-dt) = \int_{\pi/6}^{\pi/3} \frac{\cos^\alpha t}{\sin^\alpha t + \cos^\alpha t}\, dt$$

Adding this to the original: $$2g(\alpha) = \int_{\pi/6}^{\pi/3} \frac{\sin^\alpha x + \cos^\alpha x}{\cos^\alpha x + \sin^\alpha x}\, dx = \int_{\pi/6}^{\pi/3} 1\, dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$.

Therefore $$g(\alpha) = \frac{\pi}{12}$$ for all values of $$\alpha$$.

Since $$g(\alpha)$$ is constant (equal to $$\frac{\pi}{12}$$), it is an even function (a constant function satisfies $$g(\alpha) = g(-\alpha)$$).

This matches Option D: $$g(\alpha)$$ is an even function.