The integral of $\cos(3x) e^{\sin(3x)}$ can be evaluated using substitution.

Let:

$u = \sin(3x)$ Then, the derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = 3\cos(3x) \quad \Rightarrow \quad du = 3\cos(3x) dx \quad \Rightarrow \quad \frac{du}{3} = \cos(3x) dx$

Now, the integral becomes:

$\int \cos(3x) e^{\sin(3x)} dx = \frac{1}{3} \int e^u du$

The integral of $e^u$ is simply $e^u$, so:

$\frac{1}{3} \int e^u du = \frac{1}{3} e^u + C$

Finally, substituting back $u = \sin(3x)$, we get:

$\frac{1}{3} e^{\sin(3x)} + C$

So the solution is:

$\int \cos(3x) e^{\sin(3x)} dx = \frac{1}{3} e^{\sin(3x)} + C$

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

Here are 5 related questions:

1. How would you evaluate the integral of $\cos(kx) e^{\sin(kx)}$ for any constant $k$?
2. What is the derivative of $e^{\sin(3x)}$?
3. Can you compute the integral $\int e^{x^2} dx$ using similar methods?
4. How would you handle the integral $\int \sin(3x) e^{\sin(3x)} dx$?
5. What is the general rule for integrating exponential functions involving trigonometric terms?

Tip: Substitution is a powerful technique for solving integrals, especially when the integrand includes a function and its derivative!