Solving $ \int_{0}^{2\pi} e^{-x} \lvert \sin x\rvert,dx $ 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.
 A: 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
A: 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.
