The integral to be evaluated from the image is:

$\int e^{8x} \sin(e^{4x}) \, dx$

This is a challenging integral that likely requires a substitution and possibly integration by parts. Here's how to approach it:

Step 1: Substitution

Let: $u = e^{4x}$ Then, the derivative of $u$ with respect to $x$ is: $du = 4e^{4x} dx$ Or equivalently: $dx = \frac{du}{4u}$

Now, substitute into the integral: $\int e^{8x} \sin(e^{4x}) dx = \int u^2 \sin(u) \frac{du}{4u} = \frac{1}{4} \int u \sin(u) du$

Step 2: Integration by Parts

Use integration by parts, where:

• Let $v = u$, so that $dv = du$.
• Let $dw = \sin(u) du$, so that $w = -\cos(u)$.

Now, apply the integration by parts formula: $\int v dw = v w - \int w dv$ This gives: $\frac{1}{4} \int u \sin(u) du = \frac{1}{4} \left( -u \cos(u) + \int \cos(u) du \right)$ The integral of $\cos(u)$ is $\sin(u)$, so: $\frac{1}{4} \left( -u \cos(u) + \sin(u) \right)$

Step 3: Substitute Back $u = e^{4x}$

Finally, substitute back $u = e^{4x}$ into the result: $\frac{1}{4} \left( -e^{4x} \cos(e^{4x}) + \sin(e^{4x}) \right) + C$

Thus, the evaluated integral is: $\boxed{\frac{1}{4} \left( -e^{4x} \cos(e^{4x}) + \sin(e^{4x}) \right) + C}$

Would you like further clarification on any steps, or do you have any additional questions?

Here are five related questions you might explore:

1. How does the substitution method simplify more complex integrals?
2. Can you explain why integration by parts is useful for products of functions?
3. What happens when the integral involves a product of an exponential and a trigonometric function?
4. How can you apply substitution when dealing with different powers of exponential functions?
5. What are common techniques to solve integrals involving sine or cosine with exponential terms?

Tip: When facing complex integrals, always check if substitution or integration by parts can simplify the process!