How to integrate when integration by parts never ends? So I have the following integration: $\int{e^{-x}\sin{x}}$
I let u be the $e^{-x}$ and $dv = \sin{x} dx$
After doing the integration by parts I get:$(e^{-x})(-\cos{x})-\int{-e^{-x}*(-\cos{x})}$
From what it looks like, if I continue integration by parts, will it just continue going on forever?

You actually regenerate the original integral:
$$ I = \int e^{-x} \sin{x} \, dx = -e^{-x} \sin{x} + \int e^{-x} \cos{x} \, dx \\
= -e^{-x} \sin{x} -e^{-x} \cos{x} -(-1)\int e^{-x} (-\sin{x}) \, dx \\
= -e^{-x} \sin{x} -e^{-x} \cos{x} - I, $$
so
$$ I = \frac{1}{2}(-e^{-x} \sin{x} -e^{-x} \cos{x}) + C. $$
Or, you can go the other way:
$$ I = \int e^{-x} \sin{x} \, dx = e^{-x}(-\cos{x} ) - \int (-e^{-x})(-\cos{x}) \, dx \\
= -e^{-x} \cos{x} -\int e^{-x} \cos{x} \, dx \\
= -e^{-x} \cos{x} - e^{-x}\sin{x} -(-1) \int (-e^{-x}) \sin{x} \, dx \\
= -e^{-x} \cos{x} - e^{-x} \sin{x} - I
$$
