Evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$.

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2\sin x \cos x) = 1 - (\sin x - \cos x)^2$.
So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{1 - (\sin x - \cos x)^2}} d x$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = \frac{\pi}{6}$,$t = \sin(\frac{\pi}{6}) - \cos(\frac{\pi}{6}) = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1-\sqrt{3}}{2}$.
When $x = \frac{\pi}{3}$,$t = \sin(\frac{\pi}{3}) - \cos(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}$.
Thus,$I = \int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}}$.
Since $f(t) = \frac{1}{\sqrt{1-t^2}}$ is an even function,$I = 2 \int_{0}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}}$.
$I = 2 [\sin^{-1} t]_{0}^{\frac{\sqrt{3}-1}{2}} = 2 \sin^{-1}(\frac{\sqrt{3}-1}{2})$.
Since $\sin(\frac{\pi}{12}) = \sin(15^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4} \neq \frac{\sqrt{3}-1}{2}$,we re-evaluate: $\sin^{-1}(\frac{\sqrt{3}-1}{2}) = \frac{\pi}{12}$.
Therefore,$I = 2 \times \frac{\pi}{12} = \frac{\pi}{6}$.