To solve the integral of $e^{-x} \cos(9x)$, we can apply integration by parts or use a known formula for integrals involving an exponential and a trigonometric function.

The integral is: $I = \int e^{-x} \cos(9x) \, dx$

Using the formula for integrals of the form $\int e^{ax} \cos(bx) \, dx$:

The formula is: $\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}(a \cos(bx) + b \sin(bx))}{a^2 + b^2} + C$ where $a$ and $b$ are constants.

In our case:

• $a = -1$
• $b = 9$

Substituting into the formula, we get: $I = \frac{e^{-x}(-\cos(9x) + 9\sin(9x))}{(-1)^2 + 9^2} + C$ $I = \frac{e^{-x}(-\cos(9x) + 9\sin(9x))}{1 + 81} + C$ $I = \frac{e^{-x}(-\cos(9x) + 9\sin(9x))}{82} + C$

Thus, the solution is: $I = \frac{e^{-x}(-\cos(9x) + 9\sin(9x))}{82} + C$

Would you like a detailed explanation of how to derive this result?

5 Related Questions:

1. How do you apply integration by parts to similar integrals?
2. What is the general solution for integrals involving both exponential and trigonometric functions?
3. How would this solution change if the integral were $e^{2x} \cos(9x)$?
4. Can we use complex numbers to solve this type of integral?
5. What happens if we change the cosine function to sine?

Tip: When integrating functions involving both exponential and trigonometric terms, the formula method can be a quicker alternative to integration by parts.