Answer
$$\frac{{{e^{ - 3x}}\left( { - 3\sin 4x - 4\cos 4x} \right)}}{{25}} + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - 3x}}\sin 4x} dx \cr
& \cr
& {\text{Use the formula from exercise 71 on this page}} \cr
& \int {{e^{ax}}\sin bx} = \frac{{{e^{ax}}\left( {a\sin bx - b\cos bx} \right)}}{{{a^2} + {b^2}}} + C{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& \overbrace {\int {{e^{ - 3x}}\sin 4x} dx}^{\int {{e^{ax}}\sin bx} } \Rightarrow {\text{Let }}a = - 3,{\text{ }}b = 4 \cr
& {\text{Substituting into }}\left( {\bf{1}} \right) \cr
& \int {{e^{ - 3x}}\sin 4x} dx = \frac{{{e^{ - 3x}}\left( { - 3\sin 4x - 4\cos 4x} \right)}}{{{{\left( { - 3} \right)}^2} + {{\left( 4 \right)}^2}}} + C \cr
& {\text{Simplifying}} \cr
& \int {{e^{ - 3x}}\sin 4x} dx = \frac{{{e^{ - 3x}}\left( { - 3\sin 4x - 4\cos 4x} \right)}}{{25}} + C \cr} $$
