Answer
$$\frac{3}{{13}}{e^{3x}}\cos 2x + \frac{2}{{13}}{e^{3x}}\sin 2x + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{3x}}cos2x} dx \cr
& {\text{substitute }}u = \cos 2x,{\text{ }}du = - 2\sin 2xdx \cr
& dv = {e^{3x}}dx,{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& {\text{use integration by parts}} \cr
& \int {{e^{3x}}\cos 2x} dx = \frac{1}{3}{e^{3x}}\cos 2x - \int {\frac{1}{3}{e^{3x}}\left( { - 2\sin 2xdx} \right)} \cr
& \int {{e^{3x}}\cos 2x} dx = \frac{1}{3}{e^{3x}}\cos 2x + \frac{2}{3}\int {{e^{3x}}\sin 2xdx} \cr
& {\text{substitute }}u = \sin 2x,{\text{ }}du = 2\cos 2xdx \cr
& dv = {e^{3x}}dx,{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& {\text{use integration by parts}} \cr
& \int {{e^{3x}}\cos 2x} dx = \frac{1}{3}{e^{3x}}\cos 2x + \frac{2}{3}\left( {\frac{1}{3}{e^{3x}}\sin 2x - \int {\frac{1}{3}{e^{3x}}\left( {2\cos 2xdx} \right)} } \right) \cr
& \int {{e^{3x}}\cos 2x} dx = \frac{1}{3}{e^{3x}}\cos 2x + \frac{2}{3}\left( {\frac{1}{3}{e^{3x}}\sin 2x - \frac{2}{3}\int {{e^{3x}}\cos 2xdx} } \right) \cr
& \int {{e^{3x}}\cos 2x} dx = \frac{1}{3}{e^{3x}}\cos 2x + \left( {\frac{2}{9}{e^{3x}}\sin 2x - \frac{4}{9}\int {{e^{3x}}\cos 2xdx} } \right) \cr
& \int {{e^{3x}}\cos 2x} dx = \frac{1}{3}{e^{3x}}\cos 2x + \frac{2}{9}{e^{3x}}\sin 2x - \frac{4}{9}\int {{e^{3x}}\cos 2xdx} \cr
& {\text{solving for }}\int {{e^{3x}}\cos 2x} dx \cr
& \int {{e^{3x}}\cos 2x} dx + \frac{4}{9}\int {{e^{3x}}\cos 2xdx} = \frac{1}{3}{e^{3x}}\cos 2x + \frac{2}{9}{e^{3x}}\sin 2x \cr
& \frac{{13}}{9}\int {{e^{3x}}\cos 2xdx} = \frac{1}{3}{e^{3x}}\cos 2x + \frac{2}{9}{e^{3x}}\sin 2x \cr
& \int {{e^{3x}}\cos 2xdx} = \left( {\frac{9}{{13}}} \right)\frac{1}{3}{e^{3x}}\cos 2x + \left( {\frac{9}{{13}}} \right)\frac{2}{9}{e^{3x}}\sin 2x + C \cr
& \int {{e^{3x}}\cos 2xdx} = \frac{3}{{13}}{e^{3x}}\cos 2x + \frac{2}{{13}}{e^{3x}}\sin 2x + C \cr} $$