1. The problem is to evaluate the integral $$\int_0^\pi e^x \cos x \, dx$$. 2. We use integration by parts or recognize this as a standard integral of the form $$\int e^{ax} \cos(bx) \, dx$$. 3. The formula for $$\int e^{ax} \cos(bx) \, dx$$ is: $$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + C$$ 4. Here, $$a=1$$ and $$b=1$$, so: $$\int e^x \cos x \, dx = \frac{e^x}{1^2 + 1^2} (1 \cdot \cos x + 1 \cdot \sin x) + C = \frac{e^x}{2} (\cos x + \sin x) + C$$ 5. Evaluate the definite integral from 0 to $$\pi$$: $$\int_0^\pi e^x \cos x \, dx = \left[ \frac{e^x}{2} (\cos x + \sin x) \right]_0^\pi = \frac{e^\pi}{2} (\cos \pi + \sin \pi) - \frac{e^0}{2} (\cos 0 + \sin 0)$$ 6. Substitute values: $$\cos \pi = -1, \sin \pi = 0, \cos 0 = 1, \sin 0 = 0$$ 7. So: $$= \frac{e^\pi}{2} (-1 + 0) - \frac{1}{2} (1 + 0) = \frac{e^\pi}{2} (-1) - \frac{1}{2} = -\frac{e^\pi}{2} - \frac{1}{2} = -\frac{e^\pi + 1}{2}$$ Final answer: $$\boxed{-\frac{e^\pi + 1}{2}}$$