Answer
$$\frac{{2{e^{2x}}\cos 3x}}{{13}} + \frac{{3{e^{2x}}\sin 3x}}{{13}} + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{2x}}\cos 3x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral products of trigometric}} \cr
& {\text{and exponential functions}}{\text{. By the formula 43}} \cr
& \left( {43} \right):\int {{e^{au}}\cos bu} du = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\cos bu + b\sin bu} \right) + C \cr
& \cr
& {\text{with }}a = 2{\text{ and }}b = 3,\,\,\,u = x \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{{{e^{2x}}}}{{{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}}}\left( {2\cos 3x + 3\sin 3x} \right) + C \cr
& {\text{simplifying}} \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{{{e^{2x}}}}{{13}}\left( {2\cos 3x + 3\sin 3x} \right) + C \cr
& = \frac{{2{e^{2x}}\cos 3x}}{{13}} + \frac{{3{e^{2x}}\sin 3x}}{{13}} + C \cr} $$
