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

\(1\) \(\frac{2}{3}\) \(-\frac{1}{3}\) \(\frac{1}{3}\)

\(\frac{1}{3}\)

\(\begin{aligned}\text{Given, }\int \frac{dy}{y+1}&=-\int dx\frac{\cos x}{2+\sin x}\\ \log(y+1)&=-\log (2+\sin x)+c\\ \text{Setting }x&=0 \text{ and }y=1, \text{ we get }c=\log (4)\\ y+1&=\frac{4}{2+\sin x}\\ \text{Setting }x&=\frac{\pi}{2},y\left(\frac{\pi}{2}\right)=\frac{1}{3}\end{aligned}\)