To solve the integral $-\int e^{-5t} \cos(4t) \, dt$, we can use the method of integration by parts and apply it twice, or use the tabular integration method. However, an efficient approach for this specific integral is to solve it using the formula for the integral of the product of an exponential function and a trigonometric function.

The general formula for integrals of the form $\int e^{at} \cos(bt) \, dt$ is:

$\int e^{at} \cos(bt) \, dt = \frac{e^{at}}{a^2 + b^2} (a \cos(bt) + b \sin(bt)) + C$

For your integral, $a = -5$ and $b = 4$. So, let's substitute these values into the formula:

$-\int e^{-5t} \cos(4t) \, dt = -\left[\frac{e^{-5t}}{(-5)^2 + 4^2} \left(-5 \cos(4t) + 4 \sin(4t)\right)\right] + C$

First, calculate the denominator:

$(-5)^2 + 4^2 = 25 + 16 = 41$

So the integral becomes:

$-\int e^{-5t} \cos(4t) \, dt = -\left[\frac{e^{-5t}}{41} \left(-5 \cos(4t) + 4 \sin(4t)\right)\right] + C$

Simplifying further:

$-\int e^{-5t} \cos(4t) \, dt = \frac{e^{-5t}}{41} \left(5 \cos(4t) - 4 \sin(4t)\right) + C$

So the final answer is:

$\int e^{-5t} \cos(4t) \, dt = \frac{e^{-5t}}{41} \left(5 \cos(4t) - 4 \sin(4t)\right) + C$

Would you like further details or have any questions about this solution?

Here are some related questions you might find helpful:

1. How do you apply integration by parts to similar integrals?
2. What other methods can be used to solve integrals of products of exponential and trigonometric functions?
3. Can this method be used for integrals involving $\sin(bt)$ instead of $\cos(bt)$?
4. How does the sign of $a$ in the exponential function affect the integral?
5. What are some tips for recognizing when to use this formula for integration?

Tip: When dealing with the product of an exponential function and a trigonometric function, knowing the direct integration formulas can save a lot of time and effort.