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

Answer

To evaluate the integral \( \int \cos x \cos 2x \cos 5x \, dx \), we can follow these steps:

### Step 1: Multiply and Divide by 2
We start by multiplying and dividing the integral by 2:
\[ \int \cos x \cos 2x \cos 5x \, dx = \frac{1}{2} \int 2 \cos x \cos 2x \cos 5x \, dx \]

### Step 2: Use the Cosine Product Identity
Using the identity \( 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \), we can rewrite \( 2 \cos x \cos 2x \):
\[ 2 \cos x \cos 2x = \cos(3x) + \cos(-x) = \cos(3x) + \cos(x) \]
Thus, we have:
\[ \frac{1}{2} \int \left( \cos(3x) + \cos(x) \right) \cos(5x) \, dx \]

### Step 3: Expand the Integral
Now, we can expand the integral:
\[ \frac{1}{2} \int \left( \cos(3x) \cos(5x) + \cos(x) \cos(5x) \right) \, dx \]

### Step 4: Apply the Cosine Product Identity Again
For each term, we apply the cosine product identity again.

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

For \( \cos(x) \cos(5x) \):
\[ 2 \cos(x) \cos(5x) = \cos(6x) + \cos(4x) \]
So, we can rewrite the integral as:
\[ \frac{1}{4} \int \left( \cos(8x) + \cos(2x) + \cos(6x) + \cos(4x) \right) \, dx \]

### Step 5: Integrate Each Term
Now we integrate each cosine term:
\[ \int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C \]
Thus, we have:
\[ \frac{1}{4} \left( \frac{1}{8} \sin(8x) + \frac{1}{2} \sin(2x) + \frac{1}{6} \sin(6x) + \frac{1}{4} \sin(4x) \right) + C \]

### Step 6: Combine the Results
Finally, we combine the results:
\[ = \frac{1}{32} \sin(8x) + \frac{1}{8} \sin(2x) + \frac{1}{24} \sin(6x) + \frac{1}{16} \sin(4x) + C \]

### Final Answer
The evaluated integral is:
\[ \int \cos x \cos 2x \cos 5x \, dx = \frac{1}{32} \sin(8x) + \frac{1}{8} \sin(2x) + \frac{1}{24} \sin(6x) + \frac{1}{16} \sin(4x) + C \]