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

We have  
$(2+\sin x)\frac{dy}{dx}+(y+1)\cos x=0$
$\frac{dy}{y+1}=\frac{-\cos x}{2+\sin x}dx$

$\int \frac{dy}{y+1}=-\int \frac{\cos x}{2+\sin x}dx$  $\log (y+1)=-\log (2+\sin x)+\log C$
$\log (y+1)(2+\sin x)=\log C$

$(y+1)(2+\sin x)=C$
Now  $y\left(0\right)=1$
$(1+1)(2+0)=C$
$C=4$
Thus  $(y+1)(2+\sin x)=4$
Now at  $x=\frac{\pi }{2}\Rightarrow (y+1)(2+\sin \frac{\pi }{2})=4$
$(y+1)(2+1)=4$  $y+1=\frac{4}{3}$   $y=\frac{4}{3}-1=\frac{1}{3}$