$\int\limits^{\frac{\pi}{3}}_{0}\frac{sinx}{(2-cosx)^2}dx=$

$\frac{1}{2}$ 1 $\frac{2}{3}$ $\frac{1}{3}$

$\frac{1}{3}$

The correct answer is Option (4) → $\frac{1}{3}$ $\int\limits^{\frac{\pi}{3}}_{0}\frac{\sin x}{(2-\cos x)^2}dx$ Let $y=2-\cos x$ $dy=\sin xdx$ as $x→0,y→1$ $x→\frac{π}{3},y→\frac{3}{2}$ $I=\int\limits^{\frac{3}{2}}_{1}\frac{dy}{y^2}$ $=\left[\frac{-1}{y}\right]^{\frac{3}{2}}_{1}=1-\frac{2}{3}=\frac{1}{3}$