If $y=y(x)$ and $\frac{2 \sin x}{y+1} \frac{d y}{d x}=-\cos x$, $y(0)=1$, then $y(\pi / 2)$ equal

$\frac{1}{3}$

We have,

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

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

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

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

$\Rightarrow y+1=\frac{C}{2+\sin x}$

Putting $x=0$ and $y=1$ in (i), we get

$2=\frac{C}{2} \Rightarrow C=4$

Putting $C=4$ in (i), we get $y+1=\frac{4}{2+\sin x}$

Putting $x=\frac{\pi}{2}$, we get

$y+1=\frac{4}{3} \Rightarrow y=\frac{1}{3} \Rightarrow y\left(\frac{\pi}{2}\right)=\frac{1}{3}$