1. **Problem Statement:** Calculate the integral $$\int e^{-t} \cos(\omega t) \, dt$$ where $\omega$ is a constant. 2. **Formula and Approach:** We use integration by parts or recognize this as a standard integral of the form $$\int e^{at} \cos(bt) \, dt$$ which has the formula: $$\int e^{at} \cos(bt) \, dt = \frac{e^{at}}{a^2 + b^2} (a \cos(bt) + b \sin(bt)) + C$$ 3. **Apply the formula:** Here, $a = -1$ and $b = \omega$, so: $$\int e^{-t} \cos(\omega t) \, dt = \frac{e^{-t}}{(-1)^2 + \omega^2} (-1 \cdot \cos(\omega t) + \omega \sin(\omega t)) + C$$ 4. **Simplify the denominator:** $$(-1)^2 + \omega^2 = 1 + \omega^2$$ 5. **Final answer:** $$\int e^{-t} \cos(\omega t) \, dt = \frac{e^{-t}}{1 + \omega^2} (-\cos(\omega t) + \omega \sin(\omega t)) + C$$ This formula shows the integral of an exponentially decaying cosine function, combining both cosine and sine terms scaled by $\omega$ and the exponential decay factor.