Answer
$$ - \frac{1}{{17}}{e^{ - x}}sin4x - \frac{4}{{17}}{e^{ - x}}\cos 4x + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - x}}sin4x} dx \cr
& {\text{substitute }}u = sin4x,{\text{ }}du = 4\cos 4xdx \cr
& dv = {e^{ - x}}dx,{\text{ }}v = - {e^{ - x}} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int {{e^{ - x}}sin4x} dx = - {e^{ - x}}sin4x - \int {\left( { - {e^{ - x}}} \right)\left( {4\cos 4xdx} \right)} \cr
& \int {{e^{ - x}}sin4x} dx = - {e^{ - x}}sin4x + 4\int {{e^{ - x}}\cos 4xdx} \cr
& {\text{substitute }}u = \cos 4x,{\text{ }}du = - 4\sin 4xdx \cr
& dv = {e^{ - x}}dx,{\text{ }}v = - {e^{ - x}} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int {{e^{ - x}}sin4x} dx = - {e^{ - x}}sin4x + 4\left( { - {e^{ - x}}\cos 4x - \int {\left( { - {e^{ - x}}} \right)\left( { - 4\sin 4xdx} \right)} } \right) \cr
& \int {{e^{ - x}}sin4x} dx = - {e^{ - x}}sin4x + 4\left( { - {e^{ - x}}\cos 4x - 4\int {{e^{ - x}}\sin 4xdx} } \right) \cr
& \int {{e^{ - x}}sin4x} dx = - {e^{ - x}}sin4x - 4{e^{ - x}}\cos 4x - 16\int {{e^{ - x}}\sin 4xdx} \cr
& {\text{solving for }}\int {{e^{ - x}}\sin 4xdx} \cr
& \int {{e^{3x}}\cos 2x} dx + 16\int {{e^{ - x}}\sin 4xdx} = - {e^{ - x}}sin4x - 4{e^{ - x}}\cos 4x \cr
& 17\int {{e^{ - x}}\sin 4xdx} = - {e^{ - x}}sin4x - 4{e^{ - x}}\cos 4x + C \cr
& \int {{e^{ - x}}\sin 4xdx} = - \frac{1}{{17}}{e^{ - x}}sin4x - \frac{4}{{17}}{e^{ - x}}\cos 4x + C \cr} $$