Answer
\[\frac{e^{3x}}{13}\left[3\sin 2x-2\cos 2x\right]+C\]
Where $C$ ia constant of integration
Work Step by Step
Let \[I=\int e^{3x}\sin 2x\: dx\]
Using integration by parts
\[I=\sin 2x\:\int e^{3x}dx-\int \left((\sin 2x )'\:\int e^{3x}\right)dx\]
\[I=\sin 2x\left(\frac{e^{3x}}{3}\right)-\frac{2}{3}\int\cos 2x\; (e^{3x})dx\]
Again using integration by parts
\[I=\sin 2x\left(\frac{e^{3x}}{3}\right)-\frac{2}{3}\left[\cos2x\int e^{3x}dx-\int\left((\cos 2x)'\int e^{3x}\right)dx\right]\]
\[I=\sin 2x\left(\frac{e^{3x}}{3}\right)-\frac{2}{3}\left[\cos 2x \left(\frac{e^{3x}}{3}\right)+\frac{2}{3}\int \sin 2x \: e^{3x}\:dx\right]\]
\[I=\sin 2x\left(\frac{e^{3x}}{3}\right)-\frac{2}{3}\left[\cos 2x \left(\frac{e^{3x}}{3}\right)+\frac{2}{3}I\right]\]
\[I=\sin 2x\left(\frac{e^{3x}}{3}\right)-\frac{2}{9}(\cos 2x)e^{3x}-\frac{4}{9}I\]
\[\frac{13}{9}I=\left[\frac{\sin 2x}{3}-\frac{2}{9}\cos 2x\right]e^{3x}+C\]
Where $C$ is constant of integration
\[I=\frac{e^{3x}}{13}\left[3\sin 2x-2\cos 2x\right]+C\]
