how about checking by differentiating
[tex]sin(y) = \frac{A}{\sqrt{x^2+1}}[/tex]
[tex]\frac{d}{dx}sin(y) = \frac{d}{dx}\frac{A}{\sqrt{x^2+1}}[/tex]
[tex]cos(y)\frac{dy}{dx} = 2A(x^2+1)^{-3/2}\frac{-1}{2}2x[/tex]
[tex]cos(y) dy \frac{A}{\sqrt{x^2+1}} = -A(x^2+1)^{-3/2}2x.dx.sin(y)[/tex]
[tex]cos(y)dy(x^2+1)= -sin(y)2x.dx[/tex]
which is looking ok, you probably want to use your initial condition though