Chapter 8: Techniques of Integration - Section 8.6 - Integral Tables and Computer Algebra Systems - Exercises 8.6 - Page 481: 16

$$\frac{e^{-3 t}}{25}(-3 \sin 4 t-4 \cos 4 t)+C$$

Since $$\int e^{a x} \sin b x d x=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+C$$ Here $ a= -3,\ \ b=4$ \begin{align*}\int e^{-3 t} \sin 4 t d t&=\frac{e^{-3 t}}{(-3)^{2}+4^{2}}(-3 \sin 4 t-4 \cos 4 t)+C \\ &=\frac{e^{-3 t}}{25}(-3 \sin 4 t-4 \cos 4 t)+C \end{align*}