` int_(0)^(pi//6) cos x cos 3x dx`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{6}} \cos x \cos 3x \, dx \), we can follow these steps:

### Step 1: Use the product-to-sum formula
We can use the product-to-sum formula for cosine, which states:
\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
In our case, let \( A = x \) and \( B = 3x \). Thus,
\[ \cos x \cos 3x = \frac{1}{2} \left( \cos(4x) + \cos(-2x) \right) \]
Since \( \cos(-\theta) = \cos(\theta) \), we have:
\[ \cos(-2x) = \cos(2x) \]
So we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{6}} \cos x \cos 3x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{6}} \left( \cos(4x) + \cos(2x) \right) dx \]

### Step 2: Split the integral
Now we can split the integral:
\[ I = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{6}} \cos(4x) \, dx + \int_{0}^{\frac{\pi}{6}} \cos(2x) \, dx \right) \]

### Step 3: Integrate each term
Now we will integrate each term separately.

1. **Integrating \( \cos(4x) \)**:
\[ \int \cos(4x) \, dx = \frac{1}{4} \sin(4x) \]
Therefore,
\[ \int_{0}^{\frac{\pi}{6}} \cos(4x) \, dx = \left[ \frac{1}{4} \sin(4x) \right]_{0}^{\frac{\pi}{6}} = \frac{1}{4} \sin\left(4 \cdot \frac{\pi}{6}\right) - \frac{1}{4} \sin(0) = \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) \]

2. **Integrating \( \cos(2x) \)**:
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]
Therefore,
\[ \int_{0}^{\frac{\pi}{6}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{6}} = \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{6}\right) - \frac{1}{2} \sin(0) = \frac{1}{2} \sin\left(\frac{\pi}{3}\right) \]

### Step 4: Evaluate the sine values
Now we need to evaluate the sine values:
- \( \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)
- \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)

### Step 5: Substitute back into the integral
Now substituting back into our integral:
\[ I = \frac{1}{2} \left( \frac{1}{4} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right) \]
\[ = \frac{1}{2} \left( \frac{\sqrt{3}}{8} + \frac{\sqrt{3}}{4} \right) = \frac{1}{2} \left( \frac{\sqrt{3}}{8} + \frac{2\sqrt{3}}{8} \right) = \frac{1}{2} \cdot \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{16} \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{3\sqrt{3}}{16} \]