You've done the hard part of the integration by parts right; you're just off by a constant multiple. If we let $u = 4x,\; dv=\sin(2x)dx \implies du = 4\,dx,\;v=\frac{-\cos(2x)}{2}$.
When we put this together: $$\begin{align}uv - \int v\,du &= -2x\cos(2x) + \int\left(\frac{\cos(2x)}{2}\right)4\,dx\\\ &=-2x\cos(2x) + \int2\cos(2x)\,dx \end{align}$$ So, that's basically where you're at (except for the $2$ instead of the $4$). At this point, we can use $u$ substitution.
**Hint:**
> Let $u = 2x$.