If $(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$ and $y(0)=1$, then $y\left(\frac{\pi}{2}\right)$ is equal to

$\frac{1}{3}$

The given differential equation is

$(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$

$\Rightarrow (2+\sin x) d y+(y+1) \cos x d x=0$

$\Rightarrow \frac{1}{y+1} d y+\frac{\cos x}{2+\sin x} d x=0$

Integrating both sides, we get

$\log (y+1)+\log (2+\sin x)=\log C$

$\Rightarrow (y+1)(2+\sin x)=C$               .....(i)

It is given that $y(0)=1$ i.e. $y=1$ when $x=0$. Putting $x=0, y=1$ in (i), we get

$2(2+0)=C \Rightarrow C=4$

Putting $C=4$ in (i), we obtain

$(y+1)(2+\sin x)=4$          ......(ii)

Putting $x=\frac{\pi}{2}$ in (ii), we get $y=\frac{1}{3}$.