## Calculus (3rd Edition)

We have $$y'=\sec y \Longrightarrow \frac{dy}{\sec y}=dx\Longrightarrow \cos y dy=dx.$$ Now, by integration we get $$\sin y =x+c.$$ When $x=0$, $y=0$, and the constant $c=0$. Hence, $$\sin y =x\Longrightarrow y=\sin^{-1}x.$$ This completes the proof.