Answer
$$ - \frac{1}{3}\ln \left| {2 + \cos 3x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 3x}}{{2 + \cos 3x}}} dx \cr
& {\text{substitute }}u = 2 + \cos 3x,{\text{ }}du = - 3\sin 3xdx \cr
& \int {\frac{{\sin 3x}}{{2 + \cos 3x}}} dx \cr
& \int {\frac{{\left( { - 1/3} \right)du}}{u}} \cr
& = - \frac{1}{3}\int {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}}{\text{,}} \cr
& = - \frac{1}{3}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{3}\ln \left| {2 + \cos 3x} \right| + C \cr} $$
