If $y=y\left(x\right)$ and $\frac{2+\sin x}{y+1}\left(\frac{dy}{dx}\right)=-\cos x,$ $y\left(0\right)=1$ then $y\left(\frac{\pi }{2}\right)$ equals

$\frac{1}{y+1}dy=-\frac{\cos x}{2+\sin x}dx$
 Integrating, we get
 $\log (y+1)+\log k+\log (2+\sin x)=0$

$\therefore k(y+1)(2+\sin x)=1$ 
when $x=0,y=1$ where k is constant
 $\therefore 4k=1\text{ or }k=\frac{1}{4}$

$$\begin{gathered} \therefore (y+1)(2+\sin x)=4 \\ \text{ Now, Put }x=\frac{\pi }{2} \\ \therefore (y+1)3=4 \\ \therefore y=\frac{1}{3} \end{gathered}$$