No, I don't think your way works. Before your substitution, you have x in the numerator so it doesn't look like the derivative of an arcsin. After your substitution, you have $\displaystyle \sqrt{v}$ in the denominator, so again it doesn't look like the derivative of an arcsin.
But the integral is easy enough to do - you just substitute $\displaystyle w=9-(x-3)^2$ and it becomes a power of w.
- Hollywood
Your substitution looks ok, but you should have ended up with $\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int{\frac{dv}{\sqrt{ 9 -v}}} \end{align*}$. There is NO point in bringing in extra square roots, as they would require more substitutions to be able to integrate.
What you should do is let $\displaystyle \displaystyle \begin{align*} w = 9 - v \implies dw = -dv \end{align*}$ and your integral becomes $\displaystyle \displaystyle \begin{align*} -\frac{1}{2}\int{\frac{dw}{\sqrt{w}}} = -\frac{1}{2}\int{ w^{ -\frac{1}{2} } } \end{align*}$. Surely you can integrate that...
What hollywood and Prove It are saying is if it were simply $\displaystyle \sqrt{a^2 - u^2}$ then we can simply integrate. But your form of $\displaystyle \sqrt{a^2 - \left ( \sqrt{u} \right )^2$ requires another substitution for the square root over u. It's the inner square root that is the problem here, not the outer.
-Dan