In class we showed that $$ \int_{x_0}^{x(t)} \frac{dx}{\left(A^2 - x^2 \right)^{1/2}} = \pm \omega \int_0^t dt \Rightarrow \cos^{-1} \left(\frac{x}{A} \right) - \cos^{-1} \left(\frac{x_0}{A} \right) = \mp \omega t $$ will lead to the solution $$ x(t) = A \cos \left(\omega t + \phi \right)$$ where $\phi = \mp \cos^{-1} \left(\frac{x_0}{A} \right)$ with $x_0$ being the initial position.

Show that $$v(t) = -\omega A \sin \left(\omega t + \phi \right)$$ where $\phi = - \tan^{-1} \left(\frac{v_0}{\omega x_0} \right)$ assuming that the initial velocity at $t = 0 $ is $v_0$