Answer
$$\int e^{-3t}\sin4tdt=-\frac{e^{-3t}}{25}(3\sin4t+4\cos4t)+C$$
Work Step by Step
$$A=\int e^{-3t}\sin4tdt$$
Use Formula 115, which states that $$\int e^{ax}\sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C$$
for $a=-3$ and $b=4$ here:
$$A=\frac{e^{-3t}}{(-3)^2+4^2}(-3\sin4t-4\cos4t)+C$$ $$A=-\frac{e^{-3t}}{25}(3\sin4t+4\cos4t)+C$$
