Answer
$$\frac{1}{5}[e^{x} \sin 2 x-2e^{x} \cos 2 x]+c$$
Work Step by Step
Given$$\int \sin 2 x e^{x} d x $$
Let
\begin{align*}
u&=\sin 2x\ \ \ \ \ \ \ \ \ dv= e^xdx\\
du&=2\cos 2x\ \ \ \ \ \ \ v= e^x
\end{align*}
Then
\begin{aligned} \int \sin 2 x e^{x} d x &=(\sin 2 x)\left(e^{x}\right)-\int e^{x} 2 \cos 2 x d x \\ &=e^{x} \sin 2 x-2 \int \cos 2 x e^{x} d x \end{aligned}
Let
\begin{align*}
u&=\cos 2x\ \ \ \ \ \ \ \ \ dv= e^xdx\\
du&=-2\sin 2x\ \ \ \ \ \ \ v= e^x
\end{align*}
Then
\begin{aligned} \int \cos 2 x e^{x} d x &=\cos 2 x\left(e^{x}\right)-\int e^{x}(-2 \sin 2 x) d x \\ &=e^{x} \cos 2 x+2 \int e^{x} \sin 2 x d x \end{aligned}
Hence
\begin{aligned}
\int \sin 2 x e^{x} d x &=(\sin 2 x)\left(e^{x}\right)-\int e^{x} 2 \cos 2 x d x \\ &=e^{x} \sin 2 x-2e^{x} \cos 2 x-4 \int e^{x} \sin 2 x d x \\
5\int \sin 2 x e^{x} d x&=e^{x} \sin 2 x-2e^{x} \cos 2 x+c\\
\int \sin 2 x e^{x} d x&=\frac{1}{5}[e^{x} \sin 2 x-2e^{x} \cos 2 x]+c
\end{aligned}
