$\int\limits^{\frac{\pi}{3}}_{0}\frac{sinx}{(2-cosx)^2}dx=$
Answer & explanation
Correct answer: option 4
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 $x→0,y→1$
$x→\frac{π}{3},y→\frac{3}{2}$
$I=\int\limits^{\frac{3}{2}}_{1}\frac{dy}{y^2}$
$=\left[\frac{-1}{y}\right]^{\frac{3}{2}}_{1}=1-\frac{2}{3}=\frac{1}{3}$