Official solution
\[\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0\]
\[ \Rightarrow \] \[{d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0\]
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 \[y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1\]
= \[{4 \over 3} - 1\] = \[{1 \over 3}\]