To integrate the function $e^x \cos(x)$, we will employ the technique known as integration by parts.

Integration by parts is a method used for integrating the product of two functions. It involves selecting one function to differentiate and another to integrate. In this case, we will let $u = \cos(x)$ (the function to differentiate) and $dv = e^x \, dx$ (the function to integrate).

We first compute the derivatives and integrals:

• The derivative of $u$ is given by: $du = -\sin(x) \, dx$
• The integral of $dv$ yields: $v = e^x$

Applying the integration by parts formula, which states that: $\int u \, dv = uv - \int v \, du,$ we substitute our choices for $u$, $du$, $v$, and $dv$:

$\int e^x \cos(x) \, dx = e^x \cos(x) - \int e^x (-\sin(x)) \, dx$

This simplifies to:

$\int e^x \cos(x) \, dx = e^x \cos(x) + \int e^x \sin(x) \, dx$

Next, we need to integrate $e^x \sin(x)$. We can again use integration by parts, this time setting:

• $u = \sin(x)$, so $du = \cos(x) \, dx$,
• $dv = e^x \, dx$, leading to $v = e^x$.

Using the integration by parts formula again, we have:

$\int e^x \sin(x) \, dx = e^x \sin(x) - \int e^x \cos(x) \, dx$

Now, substituting this back into our earlier equation gives us:

$\int e^x \cos(x) \, dx = e^x \cos(x) + e^x \sin(x) - \int e^x \cos(x) \, dx$

Rearranging terms, we obtain:

$2 \int e^x \cos(x) \, dx = e^x (\cos(x) + \sin(x))$

Dividing both sides by $2$, we find:

$\int e^x \cos(x) \, dx = \frac{1}{2} e^x (\cos(x) + \sin(x)) + C$

where $C$ represents the constant of integration.

Thus, the integral of $e^x \cos(x)$ is given by:

$\int e^x \cos(x) \, dx = \frac{1}{2} e^x (\cos(x) + \sin(x)) + C$