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?

Here is a nice way to do it: by parts,
$$\int e^u\sin u\,du=e^u\sin u-\int e^u\cos u\,du\ .$$
Consider also
$$\int e^u\cos u\,du=e^u\cos u+\int e^u\sin u\,du\ .$$
Eliminating the cos integral,
$$\int e^u\sin u\,du=e^u\sin u-e^u\cos u-\int e^u\sin u\,du$$
so
$$2\int e^u\sin u\,du=e^u\sin u-e^u\cos u$$
and hence
$$\int e^u\sin u\,du=\frac12(e^u\sin u-e^u\cos u)\ \ldots$$
. . . not forgetting plus $C$ of course ;-)
