Answer
$\dfrac{\pi}{6}$
Work Step by Step
Plug $x =3 \sin \theta \implies dx=3 \cos \theta d \theta $
Then
$\int \dfrac{3 \cos \theta d \theta }{\sqrt{9-9 \sin^2 \theta}} = \int \dfrac{3 \cos \theta d \theta }{\sqrt{9 \cos^2 \theta}}=\int d \theta$
and
$\int_{0}^{3/2} d \theta=[sin^{-1} (x/3)]_{0}^{3/2}$
Thus, $\int \dfrac{3 \cos \theta d \theta }{\sqrt{9-9 \sin^2 \theta}} =\dfrac{\pi}{6}$
