Answer
$$\frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x + 2\cos 3x} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {{e^{2x}}\cos 3x} dx \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {e^{2x}},\,\,\,\,du = 2{e^{2x}}dx,\,\,\,\,\,dv = \cos 3xdx,\,\,\,\,v = \frac{1}{3}\sin 3x \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x - \int {\left( {\frac{1}{3}\sin 3x} \right)\left( {2{e^{2x}}} \right)dx} \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x - \frac{2}{3}\int {{e^{2x}}\sin 3xdx} \cr
& \cr
& {\text{Integration by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = {e^{2x}},\,\,\,\,du = 2{e^{2x}}dx,\,\,\,\,\,dv = \sin 3xdx,\,\,\,\,v = - \frac{1}{3}\cos 3x \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x - \frac{2}{3}\left( { - \frac{1}{3}{e^{2x}}\cos 3x - \int {\left( { - \frac{1}{3}\cos 3x} \right)\left( {2{e^{2x}}dx} \right)} } \right) \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x - \frac{2}{3}\left( { - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}\int {{e^{2x}}\cos 3xdx} } \right) \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x - \frac{4}{9}\int {{e^{2x}}\cos 3xdx} \cr
& {\text{The unknown integral now appears on both sides of the equation}}{\text{. }} \cr
& {\text{Adding the integral }}\frac{4}{9}\int {{e^{2x}}\cos 3x} dx{\text{ to both sides and adding }} \cr
& {\text{the constant }}C{\text{ we obtain}} \cr
& \int {{e^{2x}}\cos 3x} dx + \frac{4}{9}\int {{e^{2x}}\cos 3xdx} = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x + C \cr
& \frac{{13}}{9}\int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x + C \cr
& {\text{Divide both sides by }}\frac{9}{{13}} \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{9}{{13}}\left( {\frac{1}{3}{e^{2x}}\sin 3x} \right) + \frac{9}{{13}}\left( {\frac{2}{9}{e^{2x}}\cos 3x} \right) + C \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{3}{{13}}{e^{2x}}\sin 3x + \frac{2}{{13}}{e^{2x}}\cos 3x + C \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x + 2\cos 3x} \right) + C \cr} $$
