We are asked to evaluate the integral of $\frac{\sin(4x)}{16 - \cos^2(4x)} \, dx$. Let's proceed step by step.

Step 1: Recognizing a useful substitution

First, observe that the expression in the denominator has a quadratic form involving $\cos(4x)$. To simplify it, a good strategy is to try the substitution: $u = \cos(4x)$ Taking the derivative of $u$ with respect to $x$: $du = -4\sin(4x) \, dx \quad \Rightarrow \quad \sin(4x) \, dx = -\frac{1}{4} \, du$

Step 2: Substituting into the integral

Now substitute $u = \cos(4x)$ and $\sin(4x) \, dx = -\frac{1}{4} \, du$ into the integral: $\int \frac{\sin(4x)}{16 - \cos^2(4x)} \, dx = -\frac{1}{4} \int \frac{du}{16 - u^2}$

Step 3: Recognizing a standard form

The integral $\int \frac{du}{16 - u^2}$ is of the form: $\int \frac{1}{a^2 - u^2} \, du = \frac{1}{2a} \ln\left| \frac{a + u}{a - u} \right| + C$ with $a = 4$. Applying this formula:

$\int \frac{du}{16 - u^2} = \frac{1}{8} \ln\left| \frac{4 + u}{4 - u} \right| + C$

Step 4: Substituting back for $u = \cos(4x)$

Now, substitute $u = \cos(4x)$ back into the result: $-\frac{1}{4} \cdot \frac{1}{8} \ln\left| \frac{4 + \cos(4x)}{4 - \cos(4x)} \right| + C = -\frac{1}{32} \ln\left| \frac{4 + \cos(4x)}{4 - \cos(4x)} \right| + C$

Final Answer:

Thus, the integral is: $\int \frac{\sin(4x)}{16 - \cos^2(4x)} \, dx = -\frac{1}{32} \ln\left| \frac{4 + \cos(4x)}{4 - \cos(4x)} \right| + C$

Would you like further details or clarifications?

Here are 5 related questions you can explore:

1. How do you integrate functions involving $\sin(x)$ and $\cos(x)$?
2. What is the integral of $\frac{1}{a^2 - x^2}$?
3. Can you apply trigonometric identities to simplify integrals?
4. How does substitution help simplify integrals?
5. How can partial fractions be used in integration?

Tip: Always look for a substitution that simplifies the integrand, especially when dealing with trigonometric functions!