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?

If you continue (do integration by part for $\int e^{-x}\cos x\,dx$) you will get $\int e^{-x}\sin x \,dx= f(x) - \int e^{-x}\sin x\, dx$, therefore $\int e^{-x}\sin x\, dx= \frac12 f(x)$.
Every time you integrate by part you will get an extra minus, but you integrating $\sin x$ twice get one minus, that's why in this case doing integration twice works.
Edit: You also made a mistake. (Editted) $\int udv = uv -\int vdu$, so you should get
$$\int e^{-x} \sin x \,dx = -e^{-x}\cos x - \int e^{-x} \cos x \, dx$$
