How do I find the antiderivative of $sin(u)e^u$? So, I have the following indefinite integral:
$$\int sin(u)e^u \ du$$
I tried solving it using integration by parts, but it just kept repeating itself. What should I do?

You have the answer without knowing it, I think! You noted that it repeats itself (which it does). Let's see what happens. Let's let $x = \sin u$ and $dv = e^u\,du$, then $dx = \cos u \,du$ and $v = e^u$. Integration by parts gives us
$$\int\sin u e^u\,du = \sin u e^u - \int \cos u e^u\,du$$
Well we traded one integral for another. Let's try doing integration by parts again. We don't want to let $x = e^u$ (and subsequently $dv = \cos u\,du$) since this undoes our previous integration by parts. That leaves us with the case that $x = \cos u$ and $dv = e^u \,du$ to give that $dx = -\sin u\,du$ and $v = e^u$. Hence
$$\int\sin u e^u \,du = \sin u e^u - \left(\cos u e^u - \int (-\sin u) e^u \,du\right)$$
Simplifying, the right hand side becomes 
$$\sin u e^u - \cos u e^u - \int \sin u e^u \,du.$$
Note that $\int \sin u e^u \,du$ appears on both sides of the equation.. Can you take it from here?
