You are doing just fine. You completed the square, which is the key step, and ended up with $\int \frac{dx}{\sqrt{4-(x-3)^2}}.$
Now you have a formula that you want to use. It may be better not to try to remember too many formulas. We will instead find a substitution that gives us a very familiar integral.
The above integral looks kind of like $\int \frac{dw}{\sqrt{1-w^2}}.$ We look for a substitution that brings out the resemblance. Suppose that we let $x-3=2u$. Then the bottom will look like $\sqrt{4-4u^2}$, which is good, because we can then "take the $4$'s out."
So let's do it. Let $x-3=2u$, or if you prefer, let $(x-3)/2=u$. Then $dx=2\,du$ and our integral becomes $\int\frac{2u}{\sqrt{4-4u^2}}.$
But $\sqrt{4-4u^2}=2\sqrt{1-u^2}$. So our integral simplifies to $\int \frac{du}{\sqrt{1-u^2}}=\arcsin u +C.$ Finally, replace $u$ by $(x-3)/2$.
Another way to get to the right substitution is to note that $\sqrt{4-(x-3)^2}=2\sqrt{1-((x-3)/2)^2},$ which makes letting $u=(x-3)/2$ very natural.