Answer
$$ - \frac{{{e^{ - 2x}}}}{{13}}\left( {2\sin 3x + 3\cos 3x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - 2x}}\sin 3x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{The integrand has a product of trigonometric and exponential functions}} \cr
& {\text{Use formula 42}} \cr
& \left( {42} \right):\,\,\,\,\,\int {{e^{au}}\sin bu} du = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\sin bu - b\cos bu} \right) + C \cr
& \int {{e^{ - 2x}}\sin 3x} dx \to \,\,\,a = - 2,\,\,\,b = 3 \cr
& {\text{By formula }}42 \cr
& \int {{e^{ - 2x}}\sin 3x} dx = \frac{{{e^{ - 2x}}}}{{{{\left( { - 2} \right)}^2} + {{\left( 3 \right)}^2}}}\left( { - 2\sin 3x - \left( 3 \right)\cos 3x} \right) + C \cr
& {\text{simplifying}} \cr
& \int {{e^{ - 2x}}\sin 3x} dx = \frac{{{e^{ - 2x}}}}{{4 + 9}}\left( { - 2\sin 3x - 3\cos 3x} \right) + C \cr
& \int {{e^{ - 2x}}\sin 3x} dx = - \frac{{{e^{ - 2x}}}}{{13}}\left( {2\sin 3x + 3\cos 3x} \right) + C \cr} $$
