Answer
$$ - \frac{1}{3}\ln \left| {1 + \cos 3\theta } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 3\theta }}{{1 + \cos 3\theta }}d\theta } ,{\text{ with }}u = 1 + \cos 3\theta ,{\text{ }}du = - 3\sin 3\theta d\theta \cr
& - \frac{1}{3}du = \sin 3\theta d\theta \cr
& {\text{using the indicated substitution}} \cr
& \int {\frac{{\sin 3\theta }}{{1 + \cos 3\theta }}d\theta } = \int {\frac{{\left( { - 1/3} \right)du}}{u}} \cr
& = - \frac{1}{3}\int {\frac{{du}}{u}} \cr
& {\text{integrating}} \cr
& = - \frac{1}{3}\ln \left| u \right| + C \cr
& {\text{Back substitute }}u = 1 + \cos 3\theta \cr
& = - \frac{1}{3}\ln \left| {1 + \cos 3\theta } \right| + C \cr} $$
