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

$\frac{1}{3}$

The correct answer is option (1) : $\frac{1}{3}$

$\frac{2sin \, x\, }{y+1}\frac{dy}{dx} = -cos \, x$

$⇒\frac{1}{y+1}dy =-\frac{cos\, x}{2+sin x}dx$

$⇒∫\frac{1}{y+1} dy = - ∫\frac{cosx}{2+sinx}dx$

$⇒kog (y+1) = - log (2+sin x) + log C$

$⇒y + 1 = \frac{C}{2+sin x}$

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

$2=\frac{C}{2} ⇒ C=4$

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

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

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