To solve the integral \( \int \cos 2x \cdot \cos 4x \cdot \cos 6x \, dx \), we can use trigonometric identities to simplify the expression. 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 apply the product-to-sum formula for cosine:
\[
2 \cos A \cos B = \cos(A + B) + \cos(A - B)
\]
We will first combine \( \cos 2x \) and \( \cos 6x \):
\[
2 \cos 2x \cos 6x = \cos(2x + 6x) + \cos(2x - 6x) = \cos 8x + \cos(-4x) = \cos 8x + \cos 4x
\]
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 the Product-to-Sum Formula Again
For the first integral \( \int \cos 8x \cos 4x \, dx \), we apply the product-to-sum formula again:
\[
2 \cos 8x \cos 4x = \cos(8x + 4x) + \cos(8x - 4x) = \cos 12x + \cos 4x
\]
Thus, we have:
\[
\int \cos 8x \cos 4x \, dx = \frac{1}{2} \int (\cos 12x + \cos 4x) \, dx
\]
### Step 5: Combine the Integrals
Now, substituting back into our expression for \( I \):
\[
I = \frac{1}{4} \int (\cos 12x + \cos 4x) \, dx + \frac{1}{2} \int \cos^2 4x \, dx
\]
### Step 6: Solve the Integrals
1. For \( \int \cos 12x \, dx \):
\[
\int \cos 12x \, dx = \frac{1}{12} \sin 12x
\]
2. For \( \int \cos 4x \, dx \):
\[
\int \cos 4x \, dx = \frac{1}{4} \sin 4x
\]
3. For \( \int \cos^2 4x \, dx \), we use the identity:
\[
\cos^2 A = \frac{1 + \cos 2A}{2}
\]
Therefore:
\[
\int \cos^2 4x \, dx = \frac{1}{2} \int (1 + \cos 8x) \, dx = \frac{1}{2} \left( x + \frac{1}{8} \sin 8x \right)
\]
### Step 7: Combine All Parts
Putting it all together:
\[
I = \frac{1}{4} \left( \frac{1}{12} \sin 12x + \frac{1}{4} \sin 4x \right) + \frac{1}{2} \left( \frac{1}{2} x + \frac{1}{16} \sin 8x \right)
\]
### Step 8: Simplify
After simplifying, we get:
\[
I = \frac{1}{48} \sin 12x + \frac{1}{16} \sin 4x + \frac{1}{4} x + C
\]
### Final Answer
Thus, the final answer is:
\[
\int \cos 2x \cdot \cos 4x \cdot \cos 6x \, dx = \frac{1}{48} \sin 12x + \frac{1}{16} \sin 4x + \frac{1}{4} x + C
\]
