Complex integral?

Relevant wiki: Euler's Formula

$\begin{aligned} I & = \int_0^\pi e^{-2ix} {\color{#3D99F6}\sin x} \ dx & \small \color{#3D99F6} \text{By Euler's formula:} \\ & = \int_0^\pi e^{-2ix} {\color{#3D99F6}\left(\frac {e^{ix}-e^{-ix}}{2i}\right)} \ dx & \small \color{#3D99F6} e^{i\theta} = \cos \theta + i\sin \theta \\ & = \frac 1{2i} \int_0^\pi \left(e^{-ix}-e^{-3ix}\right) dx \\ & = \frac 1{2i} \left[\frac {\color{#3D99F6}e^{-3ix}}{3i}-\frac {\color{#3D99F6}e^{-ix}}i \right]_0^\pi & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \frac 1{2i} \left[\frac {\color{#3D99F6}-2}{3i}-\frac {\color{#3D99F6}-2}i \right] \\ & = \boxed{-\dfrac 23} \end{aligned}$

Using Euler's formula $e^{ix} = \cos(x) + i \sin(x)$ we see that $2i \sin(x) = e^{ix} - e^{-ix} \Longrightarrow \sin(x) = -\dfrac{i}{2} (e^{ix} - e^{-ix})$ . Thus

$\displaystyle \int \sin(x) e^{-2ix} dx = -\dfrac{i}{2} \int (e^{-ix} - e^{-3ix}) dx = \dfrac{e^{-ix}}{2} - \dfrac{e^{-3ix}}{6} + C$ ,

where the substitution method was used on the exponentials. Evaluating from $0$ to $\pi$ then gives us an answer of

$\left(\dfrac{e^{-i \pi}}{2} - \dfrac{e^{-i 3\pi}}{6}\right) - \left(\dfrac{1}{2} - \dfrac{1}{6}\right) = \left(-\dfrac{1}{2} + \dfrac{1}{6}\right) - \dfrac{1}{3} = \boxed{-\dfrac{2}{3}}$ .

Using integration by parts twice, we can derive the following general formula:

$\begin{aligned} \int{\sin(x)e^{-tx}}=\frac{-e^{-tx}(\cos(x)+t\sin(x))}{1+t^2} + C. \end{aligned}$

Now, let $t=2i$ and use the equation above to get (using Euler's formula ):

$\begin{aligned} \int_{0}^{\pi}{\sin(x)e^{-2ix}}&=\frac{-e^{-2i\pi}(\cos(\pi)+2i\sin(\pi))}{1+(2i)^2} + \frac{-e^{0}(\cos(0)+2i\sin(0))}{1+(2i)^2} \\ &=\frac{-(\cos (2\pi) + i\sin(2\pi))(\cos(\pi)+2i\sin(\pi)) + \cos(0) + 2i\sin(0)}{1+(2i)t^2}\\ &=\frac{-(1 + i\cdot 0)(-1+2i\cdot 0) + 1 + 2i\cdot 0}{1+(-4)}\\ &=\boxed{-\frac{2}{3}} \end{aligned}$