Answer
See the details below.
Work Step by Step
(a) Since $ x=3\sin \theta $ then $ dx=3\cos\theta d\theta $ and hence
$$ I=\int \frac{dx}{\sqrt{9-x^2}}=\int \frac{3\cos\theta d\theta }{\sqrt{9-9\sin^2\theta}}=\int \frac{3\cos\theta d\theta }{\sqrt{9(1-\sin^2\theta)}}\\
=\int \frac{3\cos\theta d\theta }{\sqrt{9\cos^2\theta}}=\int d\theta.$$
(b) $$ I=\int d\theta +c=\theta+c=\sin^{-1}\frac{x}{3}+c. $$
