$\int \mathrm{e}^x \cos x \mathrm{~d} x=$

Evaluate the integral $\int e^x \cos x \, dx$ using integration by parts, noting that it recurs upon itself.

Let $I = \int e^x \cos x \, dx$.

Set $u = \cos x$ and $dv = e^x \, dx$, yielding $du = -\sin x \, dx$ and $v = e^x$.

Apply integration by parts:
$I = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$.

For $\int e^x \sin x \, dx$, choose $u = \sin x$ and $dv = e^x \, dx$, so $du = \cos x \, dx$ and $v = e^x$.

This gives:
$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - I$.

Substitute into the expression for $I$:
$I = e^x \cos x + (e^x \sin x - I) = e^x (\cos x + \sin x) - I$.

Solve for $I$ by adding $I$ to both sides:
$2I = e^x (\cos x + \sin x)$,
so $I = \frac{e^x (\sin x + \cos x)}{2} + c$.

The result matches option B.