Why $\int \frac{dx}{\sqrt{9-x^2}} = sin^{-1}\frac{x}{3}$? I don't understand how the two formulas are equal since the right side involves trigonometric sine that the left is devoid of.

You can use the following substitution
$$ (*) \qquad x=\dfrac{3}{2}\sin t \implies \text{ d}x = \dfrac{3}{2} \cos t \text{ d}t $$
This'll yield the following equality
$$ \underbrace{\int \dfrac{1}{\sqrt{9-4x^2}} \text{ d}x \ \overset{(*)}= \ \dfrac{1}{3} \int \dfrac{1}{|\cos t|} \dfrac{3}{2} \cos t \text{ d}t}_{\because \ \sqrt{9-4\left(\frac{9}{4}\sin (t)\right)^2} \ = \ 3 |\cos (t)|} \ = \underbrace{\dfrac{1}{2} \int \mathrm{sgn}\left(\cos(t)\right) \text{ d}t}_{\because \ \mathrm{sgn}\left(\cos(t)\right) \ \cdot \ |\cos(t)| \ =: \ \cos(t) }  = \dfrac{1}{2} \ t \ \mathrm{sgn}\left(\cos(t)\right) + \mathcal{C} $$
Finally, reverting our substitution back, we have that
$$ t = \arcsin \left( \dfrac{2x}{3} \right) $$
and that
$$ \mathrm{sgn}\left(\cos(t)\right) = \mathrm{sgn}\left( \sqrt{1-\left(\dfrac{2x}{3}\right)^2} \right) = +1 $$
so we finally have that
$$ \int \dfrac{1}{\sqrt{9-4x^2}} \text{ d}x \ = \ \dfrac{1}{2} \arcsin \left( \dfrac{2x}{3} \right) + \mathcal{C} $$
