​$$\int_{0}^{\frac{\pi}{3}} \frac{\sin x}{(2 - \cos x)^2} \, dx =$$​​

Correct option is D

​$$\begin{aligned}&\text{We have} \\&\int_{0}^{\frac{\pi}{3}} \frac{\sin x}{(2 - \cos x)^2} \, dx \\&\text{Put } 2 - \cos x = t \Rightarrow \sin x \, dx = dt \\&\text{If } x = 0, \text{ then } t = 1 \\&\text{If } x = \frac{\pi}{3}, \text{ then } t = \frac{3}{2} \\&\int_{1}^{\frac{3}{2}} \frac{dt}{t^2} = \left[-\frac{1}{t}\right]_1^{\frac{3}{2}} = -\frac{2}{3} + 1 = \frac{1}{3}\end{aligned}$$​​