To integrate $e^{-5x} \cos(2x) \, dx$, we can use the method of integration by parts or apply a formula for integrating products of exponentials and trigonometric functions. A direct way is to use the formula for integrals of the form:

$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + C$

Here, $a = -5$ and $b = 2$. Plugging these values into the formula:

$\int e^{-5x} \cos(2x) \, dx = \frac{e^{-5x}}{(-5)^2 + 2^2} \left( -5 \cos(2x) + 2 \sin(2x) \right) + C$

Simplifying:

$\int e^{-5x} \cos(2x) \, dx = \frac{e^{-5x}}{25 + 4} \left( -5 \cos(2x) + 2 \sin(2x) \right) + C$

$\int e^{-5x} \cos(2x) \, dx = \frac{e^{-5x}}{29} \left( -5 \cos(2x) + 2 \sin(2x) \right) + C$

This is the final result.

Would you like more details on integration by parts or this formula? Here are 5 related questions to expand on this:

1. How do you apply integration by parts to this problem?
2. What is the general solution to the integral of an exponential times a sine function?
3. How would this change if the integral was $e^{-5x} \sin(2x) \, dx$?
4. What is the physical significance of this integral in engineering applications?
5. Can you solve this problem using a different method?

Tip: Knowing standard integrals can save time and simplify complex calculations!