Calculus (3rd Edition)

(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) $$I=\int d\theta +c=\theta+c=\sin^{-1}\frac{x}{3}+c.$$