The value of $∫\frac{cos\, x\, dx}{(1+sinx)(2+sinx)}$ is :

$log\left|\frac{1+sinx}{2+sinx}\right|+C, $ where C is a constant

The correct answer is Option (4) → $log\left|\frac{1+sinx}{2+sinx}\right|+C, $ where C is a constant

$∫\frac{\cos xdx}{(1+\sin x)(2+\sin x)}$

let $y=\sin x⇒dy=\cos xdx$

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

$=\int\frac{1}{1+y}-\frac{1}{2+y}dy=\log|1+y|-\log|2+y|+c$

$=\log\left|\frac{1+\sin x}{2+\sin x}\right|+c$