Answer
$\sin ^{-1} (\dfrac{x}{3})+c$
Work Step by Step
Consider the integral $\int \dfrac{dx}{\sqrt {(9-x^2)}}$
Plug $x= 3 \sin \theta \implies dx= 3 \cos \theta d \theta$ and $\sqrt {9-x^2}=3 \cos \theta$
This implies that $\int\dfrac{3 \cos \theta d \theta}{3 \cos \theta}=\int d\theta$
Also, we have $x= 3 \sin \theta \implies \theta= \sin ^{-1} (\dfrac{x}{3})$
or, $\theta+c=\sin ^{-1} (\dfrac{x}{3})+c$
