Answer
$\frac{1}{{10}}{e^{3x}}\sin x + \frac{3}{{10}}{e^{3x}}\cos x + C$
Work Step by Step
$$\eqalign{
& \int {{e^{3x}}\cos x} dx \cr
& \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {e^{3x}},{\text{ }}du = 3{e^{3x}}dx \cr
& dv = \cos xdx,{\text{ }}v = \sin x \cr
& {\text{Using the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {{e^{3x}}\cos x} dx = \left( {{e^{3x}}} \right)\left( {\sin x} \right) - \int {\left( {\sin x} \right)\left( {3{e^{3x}}} \right)dx} \cr
& \int {{e^{3x}}\cos x} dx = {e^{3x}}\sin x - 3\int {{e^{3x}}\sin xdx} \cr
& \cr
& {\text{Integrate by parts }}\int {{e^{3x}}\sin xdx} \cr
& {\text{Let }}u = {e^{3x}},{\text{ }}du = 3{e^{3x}}dx \cr
& dv = \sin xdx,{\text{ }}v = - \cos x \cr
& \int {{e^{3x}}\cos x} dx = {e^{3x}}\sin x - 3\left( {\int {{e^{3x}}\sin xdu} } \right) \cr
& \int {{e^{3x}}\cos x} dx = {e^{3x}}\sin x - 3\left( { - {e^{3x}}\cos x - \int {\left( { - \cos x} \right)\left( {3{e^{3x}}} \right)dx} } \right) \cr
& \cr
& {\text{Multiply}} \cr
& \int {{e^{3x}}\cos x} dx = {e^{3x}}\sin x + 3{e^{3x}}\cos x + 3\int {\left( { - \cos x} \right)\left( {3{e^{3x}}} \right)dx} \cr
& \int {{e^{3x}}\cos x} dx = {e^{3x}}\sin x + 3{e^{3x}}\cos x - 9\int {{e^{3x}}\cos xdx} \cr
& \cr
& {\text{Add }}9\int {{e^{3x}}\cos xdx} {\text{ to both sides}} \cr
& \int {{e^{3x}}\cos x} dx + 9\int {{e^{3x}}\cos xdx} = {e^{3x}}\sin x + 3{e^{3x}}\cos x \cr
& 10\int {{e^{3x}}\cos xdx} = {e^{3x}}\sin x + 3{e^{3x}}\cos x \cr
& \cr
& {\text{Divide both sides by 10}} \cr
& \int {{e^{3x}}\cos xdx} = \frac{1}{{10}}{e^{3x}}\sin x + \frac{3}{{10}}{e^{3x}}\cos x + C \cr} $$
