This integral can be solved in steps to match the provided result. Here's a step-by-step approach to compute the integral:

$I = \int_0^{\frac{\pi}{6}} \cos^4(3\theta) \sin^2(6\theta) \, d\theta$

Step 1: Simplify Using Trigonometric Identities

We will use two identities to simplify the powers of sine and cosine.

1. $\cos^2(x) = \frac{1 + \cos(2x)}{2}$
2. $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

Now, apply these identities to both $\cos^4(3\theta)$ and $\sin^2(6\theta)$.

Simplifying $\cos^4(3\theta)$:

$\cos^4(3\theta) = \left( \cos^2(3\theta) \right)^2 = \left( \frac{1 + \cos(6\theta)}{2} \right)^2$ $\cos^4(3\theta) = \frac{1}{4} \left( 1 + 2\cos(6\theta) + \cos^2(6\theta) \right)$ Now, simplify $\cos^2(6\theta)$: $\cos^2(6\theta) = \frac{1 + \cos(12\theta)}{2}$ So, $\cos^4(3\theta) = \frac{1}{4} \left( 1 + 2\cos(6\theta) + \frac{1 + \cos(12\theta)}{2} \right)$ Simplifying this further: $\cos^4(3\theta) = \frac{1}{4} \left( \frac{3}{2} + 2\cos(6\theta) + \frac{\cos(12\theta)}{2} \right)$

Simplifying $\sin^2(6\theta)$:

$\sin^2(6\theta) = \frac{1 - \cos(12\theta)}{2}$

Step 2: Expand the Product

Now, multiply $\cos^4(3\theta)$ by $\sin^2(6\theta)$: $\cos^4(3\theta) \sin^2(6\theta) = \left( \frac{1}{4} \left( \frac{3}{2} + 2\cos(6\theta) + \frac{\cos(12\theta)}{2} \right) \right) \cdot \frac{1 - \cos(12\theta)}{2}$ This product will give several terms which can be integrated term by term.

Step 3: Integrate Term by Term

Each term in the expansion can now be integrated over $\theta$ from $0$ to $\frac{\pi}{6}$.

After performing this process (either by hand or through a more detailed symbolic approach