Answer
$$\frac{{{e^x}}}{5}\left( {\cos 2x + 2\sin 2x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{e^x}\cos 2x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{The integrand has a product of trigonometric and exponential functions}} \cr
& {\text{Use formula 43}} \cr
& \left( {43} \right):\,\,\,\,\,\int {{e^{au}}\sin bu} du = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\cos bu + b\sin bu} \right) + C \cr
& \int {{e^x}\cos 2x} dx \to \,\,\,a = 1,\,\,\,b = 2 \cr
& {\text{By formula }}42 \cr
& \int {{e^x}\cos 2x} dx = \frac{{{e^x}}}{{{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}}}\left( {1\cos 2x + 2\sin 2x} \right) + C \cr
& {\text{simplifying}} \cr
& \int {{e^x}\cos 2x} dx = \frac{{{e^x}}}{{1 + 4}}\left( {\cos 2x + 2\sin 2x} \right) + C \cr
& \int {{e^x}\cos 2x} dx = \frac{{{e^x}}}{5}\left( {\cos 2x + 2\sin 2x} \right) + C \cr} $$
