`int cos 4x cos x dx = ?`

Options

A. `1/10 sin 5x + 1/6 sin 3x + C`
B. `1/5 cos 5x - 1/3 cos3x + C`
C. `1/10 sin 5x + 1/6 sin 3x + C`
D. none of these

Explanation

To solve the integral \( \int \cos(4x) \cos(x) \, dx \), we can use the trigonometric identity for the product of cosines:

\[ \cos(a) \cos(b) = \frac{1}{2} \left( \cos(a + b) + \cos(a - b) \right) \]

### Step-by-Step Solution:

1. **Apply the Trigonometric Identity:**
Let \( a = 4x \) and \( b = x \). Then,
\[ \cos(4x) \cos(x) = \frac{1}{2} \left( \cos(4x + x) + \cos(4x - x) \right) \]
This simplifies to:
\[ \cos(4x) \cos(x) = \frac{1}{2} \left( \cos(5x) + \cos(3x) \right) \]

2. **Rewrite the Integral:**
Substitute this back into the integral:
\[ \int \cos(4x) \cos(x) \, dx = \int \frac{1}{2} \left( \cos(5x) + \cos(3x) \right) \, dx \]
This can be rewritten as:
\[ = \frac{1}{2} \int \cos(5x) \, dx + \frac{1}{2} \int \cos(3x) \, dx \]

3. **Integrate Each Term:**
- For \( \int \cos(5x) \, dx \):
\[ \int \cos(5x) \, dx = \frac{1}{5} \sin(5x) + C_1 \]
- For \( \int \cos(3x) \, dx \):
\[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) + C_2 \]

4. **Combine the Results:**
Substitute the results back into the equation:
\[ = \frac{1}{2} \left( \frac{1}{5} \sin(5x) + \frac{1}{3} \sin(3x) \right) + C \]
This simplifies to:
\[ = \frac{1}{10} \sin(5x) + \frac{1}{6} \sin(3x) + C \]

5. **Final Result:**
Therefore, the integral \( \int \cos(4x) \cos(x) \, dx \) is:
\[ \int \cos(4x) \cos(x) \, dx = \frac{1}{10} \sin(5x) + \frac{1}{6} \sin(3x) + C \]