The value of $∫\frac{cos\, x\, dx}{(1+sinx)(2+sinx)}$ is : |
$log\left|\frac{1}{(1+sinx)(2+sinx)}\right|+C, $ where C is a constant $log\left|\frac{2+cosx}{1+cosx}\right|+C, $ where C is a constant $log\left|\frac{2+sinx}{1+sinx}\right|+C, $ where C is a constant $log\left|\frac{1+sinx}{2+sinx}\right|+C, $ where C is a constant |
$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 $y=\sin x⇒dy=\cos xdx$ $⇒I=\int\frac{dy}{(1+y)(2+y)}=\int\frac{(2+y)-(1+y)}{(1+y)(2+y)}dy$ $=\int\frac{1}{1+y}-\frac{1}{2+y}dy=\log|1+y|-\log|2+y|+c$ $=\log\left|\frac{1+\sin x}{2+\sin x}\right|+c$ |