Evaluate `int cos 4x cos 7 dx`

Answer

To evaluate the integral \( \int \cos(4x) \cos(7x) \, dx \), we can use the product-to-sum identities in trigonometry. Specifically, the identity states:

\[ 2 \cos(a) \cos(b) = \cos(a + b) + \cos(a - b) \]

### Step-by-Step Solution:

1. **Apply the Product-to-Sum Identity**:
We start with the integral:
\[ \int \cos(4x) \cos(7x) \, dx \]
We can rewrite this using the product-to-sum identity:
\[ \int \cos(4x) \cos(7x) \, dx = \frac{1}{2} \int ( \cos(4x + 7x) + \cos(4x - 7x) ) \, dx \]

2. **Simplify the Argument of Cosines**:
Simplifying the arguments:
\[ \cos(4x + 7x) = \cos(11x) \quad \text{and} \quad \cos(4x - 7x) = \cos(-3x) \]
Since \( \cos(-a) = \cos(a) \), we have:
\[ \int \cos(4x) \cos(7x) \, dx = \frac{1}{2} \int ( \cos(11x) + \cos(3x) ) \, dx \]

3. **Integrate Each Term**:
Now we can integrate each term separately:
\[ \frac{1}{2} \left( \int \cos(11x) \, dx + \int \cos(3x) \, dx \right) \]

The integrals are:
\[ \int \cos(11x) \, dx = \frac{\sin(11x)}{11} + C_1 \]
\[ \int \cos(3x) \, dx = \frac{\sin(3x)}{3} + C_2 \]

4. **Combine the Results**:
Combining these results, we get:
\[ \int \cos(4x) \cos(7x) \, dx = \frac{1}{2} \left( \frac{\sin(11x)}{11} + \frac{\sin(3x)}{3} \right) + C \]
Simplifying this gives:
\[ = \frac{\sin(11x)}{22} + \frac{\sin(3x)}{6} + C \]

### Final Answer:
Thus, the final result of the integral is:
\[ \int \cos(4x) \cos(7x) \, dx = \frac{\sin(11x)}{22} + \frac{\sin(3x)}{6} + C \]