Find antiderivative to $-2\sin(2x)e^{-4x}$ I'm trying to find an antiderivative to $(-\sin(2x)e^{-4x} dx)$. I've started out by using partial integration and come this far (see the picture 

but don't know if I'm on the right track. Does anyone have any suggestions?
Thanks!
/Nick
 A: Hint
Not only you’re on the right track... but you’re almost there. Just put the integral you’re looking for on the other side of the equality.
A: Let $I$ be the integral you intend to evaluate. Notice that $I$ is in the RHS of some equality for $I$. Rearrange.
A: Had a prof tell me once, A) You do math so you can be lazy and B)You don't understand the problem unless you know five ways of solving it. 
We can be lazy and use Euler's Formula: $e^{ix}=\cos{(x)}+i\sin{(x)}$
$f(x)=-2\sin{(2x)}e^{-4x}$
$f(x)=-2e^{-4x}\frac{e^{2ix}-e^{-2ix}}{2i}=-2\frac{1}{2i}(e^{(-4+2i)x}-e^{(-4-2i)x})$
Can  you carry it from there? It is now amenable to the standard procedure for integrating exponentials. 
A: It is the imaginary part of the antiderivative of $\, \mathrm e^{2ix}\mathrm e^{4x}=-\mathrm e^{(4+2i) x}$, i.e. it is
$$-\frac 1{2(2+i)}\mathrm e^{(4+2i)x}.$$
There remains to find explicitly this imaginary part. 
Note that for the same p price you'll also get $\;\displaystyle\int (-4\cos 2x\mathrm e^{4x})\,\mathrm dx$.
