$\displaystyle \int e^x \sin 2x dx$

I used the integration by parts method.

$\displaystyle \int v\frac{du}{dx}=(uv)-\int u\frac{dv}{dx}$

I let $\displaystyle v=\sin 2x\rightarrow \frac{dv}{dx}=2\cos 2x$

and $\displaystyle \frac{du}{dx}=e^x\rightarrow u=e^x$

so,

$\displaystyle uv-\int u\frac{dv}{dx}$

$\displaystyle =e^x(\sin 2x)-\int 2e^x\cos 2x dx$

This is the part I am stuck at. If I integrate $\displaystyle 2e^x\cos 2x$ by parts I get another product. And it will continue for I don't know what because I can't get rid of the $\displaystyle e^x$.

Any help? I've definitely done something wrong. Thanks