To solve the integral

$\int_{-1}^1 e^x \cos(2x) \, dx$

we will use integration by parts or recognize that this integral involves an odd function over a symmetric interval.

Step 1: Determine the nature of the integrand

Since we are integrating from $-1$ to $1$, we should check if the function $f(x) = e^x \cos(2x)$ is odd or even. If $f(x)$ is odd, the integral over $[-1, 1]$ will be zero.

1. Check for Odd or Even Functionality:

   • A function $f(x)$ is even if $f(-x) = f(x)$.
   • A function $f(x)$ is odd if $f(-x) = -f(x)$.

2. Evaluate $f(-x)$: $f(-x) = e^{-x} \cos(-2x) = e^{-x} \cos(2x)$ Since $e^x$ and $\cos(2x)$ are neither purely even nor purely odd, we find that $f(x)$ itself is not strictly odd or even.

Step 2: Direct Integration Using By-Parts Strategy

To evaluate the integral directly, we proceed by integrating by parts or applying complex exponentials. Let's proceed with the approach using complex exponentials.

Let: $\cos(2x) = \text{Re}\left(e^{i 2x}\right)$

Then, $\int e^x \cos(2x) \, dx = \text{Re}\left(\int e^x e^{i 2x} \, dx\right) = \text{Re}\left(\int e^{(1 + 2i)x} \, dx\right)$

Now, integrating $e^{(1 + 2i)x}$: $\int e^{(1 + 2i)x} \, dx = \frac{e^{(1 + 2i)x}}{1 + 2i} + C$

Evaluating from $-1$ to $1$