$\int\limits^{\frac{\pi}{2}}_{0}\frac{cos\, x}{(1+sin\, x)(2+sin\, x)}dx=$

$log\frac{4}{3}$

The correct answer is Option (4) → $\log\frac{4}{3}$

let $y=\sin x$

$dy=\cos xdx$

as $x→0,x→\frac{π}{2}$

$y→0,y→1$

so $I=\int\limits_0^1\frac{dy}{(1+y)(2+y)}=\int\limits_0^1\frac{(2+y)-(1+y)}{(1+y)(2+y)}dy$

$=\int\limits_0^1\frac{1}{1+y}-\frac{1}{2+y}dy=\left[\log(1+y)-\log(2+y)\right]_0^1$

$=\left[\log\left|\frac{1+y}{2+y}\right|\right]_0^1=\log|\frac{2}{3}|-\log|\frac{1}{2}|$

$=\log\frac{4}{3}$