$$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$$

$$ \Rightarrow $$ $${d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0$$

On integrating, we get

(2 + sin x) (y + 1) = C

At x = 0, y = 1 we have

(2 + sin 0) (1 + 1) = C

$$ \Rightarrow $$ C = 4

$$ \Rightarrow $$ $$\left( {y + 1} \right) = {4 \over {2 + \sin x}}$$

$$ \Rightarrow $$ y = $${4 \over {2 + \sin x}} - 1$$

Now $$y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1$$

= $${4 \over 3} - 1$$ = $${1 \over 3}$$