`int cos x cos 3x dx`

Answer

To solve the integral \( \int \cos x \cos 3x \, dx \), we can use the product-to-sum identities. Here’s a step-by-step solution:

### Step 1: Use the Product-to-Sum Identity
We know that:
\[ \cos A \cos B = \frac{1}{2} \left( \cos(A + B) + \cos(A - B) \right) \]
In our case, let \( A = x \) and \( B = 3x \). Therefore:
\[ \cos x \cos 3x = \frac{1}{2} \left( \cos(x + 3x) + \cos(x - 3x) \right) = \frac{1}{2} \left( \cos(4x) + \cos(-2x) \right) \]
Since \( \cos(-\theta) = \cos(\theta) \), we can simplify this to:
\[ \cos x \cos 3x = \frac{1}{2} \left( \cos(4x) + \cos(2x) \right) \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \cos x \cos 3x \, dx = \int \frac{1}{2} \left( \cos(4x) + \cos(2x) \right) \, dx \]
This simplifies to:
\[ \frac{1}{2} \int \cos(4x) \, dx + \frac{1}{2} \int \cos(2x) \, dx \]

### Step 3: Integrate Each Term
Now we will integrate each term separately.

1. For \( \int \cos(4x) \, dx \):
\[ \int \cos(4x) \, dx = \frac{1}{4} \sin(4x) + C_1 \]

2. For \( \int \cos(2x) \, dx \):
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C_2 \]

### Step 4: Combine the Results
Now substituting back into our integral:
\[ \frac{1}{2} \left( \frac{1}{4} \sin(4x) + \frac{1}{2} \sin(2x) \right) + C \]
This simplifies to:
\[ \frac{1}{8} \sin(4x) + \frac{1}{4} \sin(2x) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \cos x \cos 3x \, dx = \frac{1}{8} \sin(4x) + \frac{1}{4} \sin(2x) + C \]

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