Evaluate `int cos x cos 2x cos 5x dx`

Answer

To evaluate the integral \( \int \cos x \cos 2x \cos 5x \, dx \), we can use the product-to-sum identities to simplify the expression. Here are the steps to solve the integral:

### Step 1: Use the Product-to-Sum Formula
We know that:
\[ 2 \cos a \cos b = \cos(a + b) + \cos(a - b) \]
We will apply this formula to \( \cos x \) and \( \cos 2x \).

### Step 2: Simplify \( \cos x \cos 2x \)
Using the formula:
\[ \cos x \cos 2x = \frac{1}{2} \left( \cos(3x) + \cos(-x) \right) = \frac{1}{2} \left( \cos(3x) + \cos x \right) \]
Thus, we can rewrite the integral as:
\[ \int \cos x \cos 2x \cos 5x \, dx = \int \left( \frac{1}{2} \left( \cos(3x) + \cos x \right) \cos 5x \right) \, dx \]

### Step 3: Distribute \( \cos 5x \)
Now we distribute \( \cos 5x \):
\[ = \frac{1}{2} \int \left( \cos(3x) \cos 5x + \cos x \cos 5x \right) \, dx \]

### Step 4: Simplify Each Term
We will simplify each term using the product-to-sum formula again.

1. For \( \cos(3x) \cos(5x) \):
\[ \cos(3x) \cos(5x) = \frac{1}{2} \left( \cos(8x) + \cos(-2x) \right) = \frac{1}{2} \left( \cos(8x) + \cos(2x) \right) \]

2. For \( \cos x \cos(5x) \):
\[ \cos x \cos(5x) = \frac{1}{2} \left( \cos(6x) + \cos(-4x) \right) = \frac{1}{2} \left( \cos(6x) + \cos(4x) \right) \]

### Step 5: Combine the Integrals
Now we can combine everything:
\[ = \frac{1}{2} \int \left( \frac{1}{2} \left( \cos(8x) + \cos(2x) \right) + \frac{1}{2} \left( \cos(6x) + \cos(4x) \right) \right) \, dx \]
\[ = \frac{1}{4} \int \left( \cos(8x) + \cos(2x) + \cos(6x) + \cos(4x) \right) \, dx \]

### Step 6: Integrate Each Cosine Term
Now we can integrate each term:
\[ = \frac{1}{4} \left( \frac{\sin(8x)}{8} + \frac{\sin(2x)}{2} + \frac{\sin(6x)}{6} + \frac{\sin(4x)}{4} \right) + C \]

### Final Result
Thus, the final result is:
\[ = \frac{1}{32} \sin(8x) + \frac{1}{8} \sin(2x) + \frac{1}{24} \sin(6x) + \frac{1}{16} \sin(4x) + C \]