What is the integral of sin4x all over 16-cos^2 4x and dx?

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!

Learn how to solve the integral of sin(4x) over (16 - cos^2(4x)) with detailed trigonometric substitution and integration techniques. This solution covers advanced calculus concepts like logarithmic integration and substitution. Suitable for college-level students.

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