Twisty by parts

Below is the solution to the extra challenging problem that was posed during tutorial:
Integrate e^x sin x with respect to x.

By parts with u = sin x and dv/dx = e^x,
so that du/dx = cos x and v = e^x.

Hence, Integral(e^x sin x)
= e^x sin x - Integral(e^x cos x)

Then, another by parts with u = cos x and dv/dx = e^x
so that du/dx = -sin x and v = e^x, and thus, we obtain

= e^x sin x - (e^x cos x - Integral(- e^x sin x))
= e^x sin x - e^x cos x - Integral(e^x sin x)

Re-arranging, we have
Integral(e^x sin x) = (e^x sin x - e^x cos x)/2 + C