Answer
\[e^{−(2+i)x}=e^{-2x}\cos x+i(-e^{-2x}\sin x)\]
Work Step by Step
We will use the Euler's formula
\[e^{i\theta}=\cos \theta+i\sin\theta\]
We can write
$\;e^{−(2+i)x}=e^{-2x-ix}$
$e^{−(2+i)x}=e^{-2x}e^{-ix}$
By using Euler's formula
$e^{−(2+i)x}=e^{-2x}\left[\cos(-x)+i\sin(-x)\right]$
$e^{−(2+i)x}=e^{-2x}\left[\cos x-i\sin x\right]$
$e^{−(2+i)x}=e^{-2x}\cos x+i(-e^{-2x}\sin x)$
Here $\;u(x)=e^{-2x}\cos x$ and $\;v(x)=-e^{-2x}\sin x$
Hence $\;\;e^{−(2+i)x}=e^{-2x}\cos x+i(-e^{-2x}\sin x)$.
