## Thomas' Calculus 13th Edition

$\dfrac{-e^{-2x}}{13}[3 \cos 3x+2 \sin 3x]$
Here, $I_1=\int e^{-2x} \sin 3x dx=\dfrac{-1}{3}(e^{-2x})( \cos 3x) - \dfrac{2}{3} \int (e^{-2x}) (\cos 3x) dx$ $\implies I_1=\dfrac{9}{4}[(-1/3)(e^{-2x})( \cos 3x) - (2/3) \int e^{-2x} (\cos 3x) dx]$ Next, let us consider that $I_2=\int e^{-2x} \sin 3x dx=-\dfrac{1}{2} (e^{-2x}) (\sin 3x) - \dfrac{3}{2} \int e^{-2x} \cos 3x dx$ Now, $I=I_1+I_2$ $I=(\dfrac{9}{4}+1) \int e^{-2x} \sin 3x dx$ $\implies I=(\dfrac{-3}{4})e^{-2x} \cos 3x-(\dfrac{1}{2})e^{-2x} \sin 3x$ Thus, $\implies I=-e^{-2x}[(3/4) \cos 3x+(1/2)\sin 3x]=\dfrac{-e^{-2x}}{13}[3 \cos 3x+2 \sin 3x]$