The integral you're referring to is:

$\int 5 \cos^3(4x) \, dx$

To solve this, we'll use a trigonometric identity to simplify the expression.

1. Start with the identity for $\cos^3(x)$:

$\cos^3(x) = \cos(x)\left(\frac{1 + \cos(2x)}{2}\right)$

Thus, the given integral becomes:

$\int 5 \cos^3(4x) \, dx = \int 5 \cos(4x) \left(\frac{1 + \cos(8x)}{2}\right) \, dx$

2. Simplify the integral:

$= \frac{5}{2} \int \cos(4x)(1 + \cos(8x)) \, dx$

Now split this into two integrals:

$= \frac{5}{2} \left( \int \cos(4x) \, dx + \int \cos(4x) \cos(8x) \, dx \right)$

3. Solve each integral separately:

For the first integral: $\int \cos(4x) \, dx = \frac{1}{4} \sin(4x)$

For the second integral, we can use the product-to-sum identity:

$\cos(A)\cos(B) = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$

Thus:

[ \cos(4x) \cos(8x) = \frac{1}{2}[\cos(-4x) + \cos(12x)] = \frac{1}{2}[\cos(4x) + \cos(12x)] ]

Now integrate both terms:

$\int \cos(4x) \cos(8x) \, dx = \frac{1}{2} \left( \int \cos(4x) \, dx + \int \cos(12x) \, dx \right)$

The integrals are straightforward:

$\int \cos(12x) \, dx = \frac{1}{12} \sin(12x)$

So now we combine everything:

$\frac{5}{2} \left( \frac{1}{4} \sin(4x) + \frac{1}{2} \left( \frac{1}{4} \sin(4x) + \frac{1}{12} \sin(12x) \right) \right)$

Simplify this expression:

$= \frac{5}{8} \sin(4x) + \frac{5}{24} \sin(12x) + C$

where $C$ is the constant of integration.

Thus, the final answer is:

$\boxed{\frac{5}{8} \sin(4x) + \frac{5}{24} \sin(12x) + C}$

---

Would you like a more detailed explanation of any step? Here are 5 related questions:

1. What is the product-to-sum identity for trigonometric functions?
2. How do you use substitution to solve integrals like $\int \cos(4x) \, dx$?
3. Why do we split the original integral using trigonometric identities?
4. What are common trigonometric identities used in integration?
5. How do you evaluate more complex trigonometric integrals?

Tip: Always check for trigonometric identities that can simplify powers of sine or cosine before integrating directly!