Answer
$$\frac{1}{3}\sqrt {5 + 2\sin 3x} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos 3x}}{{\sqrt {5 + 2\sin 3x} }}} dx \cr
& {\text{substitute }}u = 5 + 2\sin 3x,{\text{ }}du = 6\cos 3xdx \cr
& = \int {\frac{{\cos 3x}}{{\sqrt {5 + 2\sin 3x} }}} dx = \int {\frac{{\left( {1/6} \right)du}}{{\sqrt u }}} \cr
& = \frac{1}{6}\int {{u^{ - 1/2}}du} \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{6}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \frac{1}{3}{u^{1/2}} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{3}{\left( {5 + 2\sin 3x} \right)^{1/2}} + C \cr
& = \frac{1}{3}\sqrt {5 + 2\sin 3x} + C \cr} $$
