`int cos 4x. cos 2x dx`

Answer

To solve the integral \( \int \cos 4x \cdot \cos 2x \, dx \), we can use the product-to-sum identities for cosine. The relevant identity is:

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

### Step-by-step Solution:

1. **Identify A and B**:
Here, let \( A = 4x \) and \( B = 2x \).

2. **Apply the product-to-sum identity**:
Using the identity, we have:
\[ \cos 4x \cdot \cos 2x = \frac{1}{2} \left( \cos(4x + 2x) + \cos(4x - 2x) \right) \]

3. **Simplify the expression**:
This simplifies to:
\[ \cos 4x \cdot \cos 2x = \frac{1}{2} \left( \cos(6x) + \cos(2x) \right) \]

4. **Set up the integral**:
Now we can rewrite the integral:
\[ \int \cos 4x \cdot \cos 2x \, dx = \int \frac{1}{2} \left( \cos(6x) + \cos(2x) \right) \, dx \]

5. **Factor out the constant**:
Factor out \( \frac{1}{2} \):
\[ = \frac{1}{2} \int \left( \cos(6x) + \cos(2x) \right) \, dx \]

6. **Integrate each term**:
Now we can integrate each term separately:
\[ \int \cos(6x) \, dx = \frac{1}{6} \sin(6x) + C_1 \]
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C_2 \]

7. **Combine the results**:
Therefore, we have:
\[ \frac{1}{2} \left( \frac{1}{6} \sin(6x) + \frac{1}{2} \sin(2x) \right) + C \]
Simplifying this gives:
\[ = \frac{1}{12} \sin(6x) + \frac{1}{4} \sin(2x) + C \]

### Final Answer:
\[ \int \cos 4x \cdot \cos 2x \, dx = \frac{1}{12} \sin(6x) + \frac{1}{4} \sin(2x) + C \]