The integral provided in the image is:

$\int e^{-2x} \cos(5x) \, dx$

To solve this, we can use integration by parts or recognize it as a standard integral for functions involving $e^{ax} \cos(bx)$.

General Formula

The integral of $e^{ax} \cos(bx)$ is given by:

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

Here:

• $a = -2$
• $b = 5$

Substituting $a = -2$ and $b = 5$:

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

Simplifying:

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

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

Final Answer:

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

Let me know if you'd like further clarifications or more examples of this type of integral!

Related Questions:

1. How can we derive the formula for $\int e^{ax} \cos(bx) \, dx$ using integration by parts?
2. What happens if we replace $\cos(5x)$ with $\sin(5x)$?
3. Can this integral be solved using substitution techniques?
4. What is the physical interpretation of such an integral in engineering contexts?
5. How does the behavior of the function $e^{-2x} \cos(5x)$ change as $x \to \infty$?

Tip:

Always check if the integrand matches a standard integral form; this can save significant time!