Answer
$\dfrac{\pi}{6}$
Work Step by Step
Let us consider $x =3 \sin \theta $
and $dx=3 \cos \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{3 \cos \theta d \theta }{\sqrt{9-9 \sin^2 \theta}} = \int d \theta$
Now, integrate with limits:
$\int_{0}^{3/2} d \theta=[sin^{-1} (x/3)]_{0}^{3/2}=\dfrac{\pi}{6}$
