Answer
$$\frac{1}{3}{\tanh ^{ - 1}}\frac{{\sin \theta }}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \theta }}{{9 - {{\sin }^2}\theta }}} d\theta \cr
& {\text{Let }}u = \sin \theta \Rightarrow du = \cos \theta d\theta \cr
& \int {\frac{{\cos \theta }}{{9 - {{\sin }^2}\theta }}} d\theta = \int {\frac{{du}}{{9 - {u^2}}}} \cr
& {\text{Integrate by using the formula }} \cr
& \int {\frac{{du}}{{{a^2} - {u^2}}} = \frac{1}{a}{{\tanh }^{ - 1}}\frac{u}{a} + C,{\text{ for }}\left| u \right| < a} {\text{ }} \cr
& {\text{ or }}\frac{1}{a}{\coth ^{ - 1}}\frac{u}{a} + C,{\text{ for }}\left| u \right| > a \cr
& \left| {\sin \theta } \right| < 3,{\text{ then }} \cr
& \int {\frac{{du}}{{9 - {u^2}}}} = \frac{1}{a}{\tanh ^{ - 1}}\frac{u}{a} + C \cr
& {\text{Back - substitute }}u = \sin \theta {\text{ and }}a = 3 \cr
& \int {\frac{{\cos \theta }}{{9 - {{\sin }^2}\theta }}} d\theta = \frac{1}{3}{\tanh ^{ - 1}}\frac{{\sin \theta }}{3} + C \cr} $$
