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

\(\begin{array}{ll}
& (2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0 \\
\therefore \quad & \frac{1}{y+1} \mathrm{~d} y=\frac{-\cos x}{2+\sin x} \mathrm{~d} x \\
& \text { Integrating on both sides, we get } \\
& \log (y+1)=-\log (2+\sin x)+\mathrm{c} \\
& \text { When } x=0, y=1 \\
& \Rightarrow \mathrm{c}=2 \log 2=\log 4 \\
\therefore \quad & \text { (i) } \Rightarrow \log (y+1)=-\log (2+\sin x)+\log 4 \\
& \Rightarrow \log (y+1)=\log \left(\frac{4}{2+\sin x}\right)
\end{array}\)
\(\begin{aligned} & \Rightarrow y+1=\frac{4}{2+\sin x} \\ & \Rightarrow y=\frac{4}{2+\sin x}-1 \\ \therefore \quad & y\left(\frac{\pi}{2}\right)=\frac{4}{3}-1=\frac{1}{3}\end{aligned}\)