`int cos 2x . cos 4x . cos 6x dx`

Answer

To solve the integral \( \int \cos 2x \cdot \cos 4x \cdot \cos 6x \, dx \), we will use trigonometric identities and properties of integration. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \cos 2x \cdot \cos 4x \cdot \cos 6x \, dx \]

### Step 2: Use the Product-to-Sum Formula
We can use the product-to-sum identities. First, we multiply and divide by 2:

\[ I = \frac{1}{2} \int 2 \cos 2x \cdot \cos 6x \cdot \cos 4x \, dx \]

Using the identity \( 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \), we apply it to \( 2 \cos 2x \cos 6x \):

\[ 2 \cos 2x \cos 6x = \cos(8x) + \cos(-4x) = \cos(8x) + \cos(4x) \]

Thus, we can rewrite the integral as:

\[ I = \frac{1}{2} \int (\cos 8x + \cos 4x) \cos 4x \, dx \]

### Step 3: Expand the Integral
Now we expand the integral:

\[ I = \frac{1}{2} \int \cos 8x \cos 4x \, dx + \frac{1}{2} \int \cos^2 4x \, dx \]

### Step 4: Apply Product-to-Sum Formula Again
For the first integral \( \int \cos 8x \cos 4x \, dx \), we use the product-to-sum formula again:

\[ 2 \cos 8x \cos 4x = \cos(12x) + \cos(4x) \]

Thus,

\[ \int \cos 8x \cos 4x \, dx = \frac{1}{2} \int (\cos(12x) + \cos(4x)) \, dx \]

### Step 5: Integrate Each Term
Now we can integrate each term:

1. \( \int \cos(12x) \, dx = \frac{\sin(12x)}{12} \)
2. \( \int \cos(4x) \, dx = \frac{\sin(4x)}{4} \)
3. For \( \int \cos^2(4x) \, dx \), we use the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \):

\[ \int \cos^2(4x) \, dx = \frac{1}{2} \int (1 + \cos(8x)) \, dx = \frac{1}{2} \left( x + \frac{\sin(8x)}{8} \right) \]

### Step 6: Combine All Parts
Putting it all together, we have:

\[ I = \frac{1}{2} \left( \frac{1}{2} \left( \frac{\sin(12x)}{12} + \frac{\sin(4x)}{4} \right) + \frac{1}{2} \left( x + \frac{\sin(8x)}{8} \right) \right) + C \]

### Final Result
Thus, the final result of the integral is:

\[ I = \frac{\sin(12x)}{48} + \frac{\sin(4x)}{16} + \frac{x}{4} + \frac{\sin(8x)}{32} + C \]