## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{\pi}{6}$
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: $\int_{0}^{3/2} d \theta=[sin^{-1} (x/3)]_{0}^{3/2}=\dfrac{\pi}{6}$