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

We have,

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

$\text{ Putting }x=0\text{ and }y=1\text{ in (i), we get }$

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

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

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

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