Answer
$\dfrac{-\sqrt {9-w^2}}{w} -\sin^{-1} (w/3)+C$
Work Step by Step
Let us consider $w =3 \sin \theta $
and $dw=3 \cos \theta d\theta $
Now, the given integral becomes:
$\int \dfrac{\sqrt{9-9 \sin^2 \theta}}{9 \sin^2 \theta} (3 \cos \theta) d \theta = \int (\csc^2 \theta -1)d \theta$
Now, integrate and plug in $w =3 \sin \theta $:
$- \cot \theta -\theta+C=\dfrac{-\sqrt {9-w^2}}{w} -\sin^{-1} (w/3)+C$
