Since it involves an absolute value, I assume I need to split it into two cases? For $ 0 \le x \le \pi $
$$ \int_{0}^{\pi} e^{-x} \sin x\,dx $$
and for $ \pi \le x \le 2\pi $
$$ \int_{\pi}^{2\pi} e^{-x} \cdot(-\sin x)dx$$
I don't really know where to go from here. I've tried to integrate both of these, but I always end up with the wrong answer.
I want to point that the integrals
$$I = \int e^{ax}\cos{bx} \qquad J = \int e^{ax}\sin{bx}$$
can be solved by using Euler's formula, as follows
Consider the integral
$$A = \int e^{ax}\cos{bx}dx -i\int e^{ax}\sin{bx}dx = \int e^{(a+ib)x} dx$$ Therefore $I = Re\{A\}$ and $J = Im\{A\}$
This approach is suitable for this problem
Both integrals are the same except the minus sign and the end points of the intervals. Let's use integration by part to find the anti-derivative for $e^{-x}\sin x$.
$\displaystyle \int e^{-x}\sin xdx = -\displaystyle \int \sin xd(e^{-x}) = -\left(\sin xe^{-x} - \displaystyle \int e^{-x}\cos xdx\right) = -\sin xe^{-x} + \displaystyle \int e^{-x}\cos xdx = -\sin xe^{-x} - \displaystyle \int \cos xd(e^{-x}) = -\sin xe^{-x} - \cos xe^{-x} - \displaystyle \int \sin xe^{-x}dx \Rightarrow \displaystyle \int \sin xe^{-x}dx = \dfrac{-\sin xe^{-x} - \cos xe^{-x}}{2}= F(x)$.
From here, to evaluate each integral, we use the FTC: $\displaystyle \int_{0}^{\pi} e^{-x}\sin xdx = F(\pi) - F(0)$, and similarly for the second one.
