I went wrong somewhere here. First let's say:

$\displaystyle L=\int e^{-5x}sin(x)dx$

Let's integrate that by parts. Let's start with:

$\displaystyle u=e^{-5x}$

$\displaystyle u'=-5e^{-5x}$

$\displaystyle v=-cosx$

$\displaystyle v'=sinx$

Next:

$\displaystyle (e^{-5x})(-cosx-\int(-5e^{-5x})(-cosx)dx$

$\displaystyle (e^{-5x})(-cosx+5\int(e^{-5x})(-cosx)dx$

Let's integrate by parts again on that new integral because I think I can get back to the original integral:

$\displaystyle \int(e^{-x})(-cosx)dx$

$\displaystyle u=-cosx$

$\displaystyle u'=sinx$

$\displaystyle v=-\frac{1}{5}e^{-5x}$

$\displaystyle v'=e^{-5x}$

So we have:

$\displaystyle (-cosx)(-\frac{1}{5}e^{-5x})-\int(sinx)(-\frac{1}{5}e^{-5x})$

$\displaystyle (-cosx)(-\frac{1}{5}e^{-5x})+\frac{1}{5}\int(e^{-5x})(sinx)$

All together:

$\displaystyle (e^{-5x})(-cosx)+5\left [(-cosx)(-\frac{1}{5}e^{-5x})+\frac{1}{5}L\right ]=L$

I don't think I did it right because the my next step appears to be a dead end:

$\displaystyle -e^{-5x}cosx+e^{-5x}+L=L$

I did something wrong with the coefficients or the signs methinks... or everything. That's always a possibility with me.