`int" cos"^(4) " 2x dx "`

Answer

To solve the integral \( \int \cos^4(2x) \, dx \), we can follow these steps:

### Step 1: Use the identity for \( \cos^2(x) \)

We know that:
\[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \]
Thus, we can express \( \cos^4(2x) \) as:
\[ \cos^4(2x) = \left(\cos^2(2x)\right)^2 = \left(\frac{1 + \cos(4x)}{2}\right)^2 \]

### Step 2: Expand the square

Now, we expand \( \left(\frac{1 + \cos(4x)}{2}\right)^2 \):
\[ \cos^4(2x) = \frac{(1 + \cos(4x))^2}{4} = \frac{1 + 2\cos(4x) + \cos^2(4x)}{4} \]

### Step 3: Substitute \( \cos^2(4x) \)

Next, we apply the identity for \( \cos^2(x) \) again to \( \cos^2(4x) \):
\[ \cos^2(4x) = \frac{1 + \cos(8x)}{2} \]
Substituting this into our expression gives:
\[ \cos^4(2x) = \frac{1 + 2\cos(4x) + \frac{1 + \cos(8x)}{2}}{4} \]
This simplifies to:
\[ \cos^4(2x) = \frac{1 + 2\cos(4x) + \frac{1}{2} + \frac{\cos(8x)}{2}}{4} = \frac{\frac{3}{2} + 2\cos(4x) + \frac{\cos(8x)}{2}}{4} \]
\[ = \frac{3 + 4\cos(4x) + \cos(8x)}{8} \]

### Step 4: Integrate term by term

Now we can integrate:
\[ \int \cos^4(2x) \, dx = \int \frac{3 + 4\cos(4x) + \cos(8x)}{8} \, dx \]
This can be split into separate integrals:
\[ = \frac{1}{8} \int (3 + 4\cos(4x) + \cos(8x)) \, dx \]
\[ = \frac{1}{8} \left( \int 3 \, dx + 4 \int \cos(4x) \, dx + \int \cos(8x) \, dx \right) \]

### Step 5: Calculate each integral

1. \( \int 3 \, dx = 3x \)
2. \( \int \cos(4x) \, dx = \frac{\sin(4x)}{4} \)
3. \( \int \cos(8x) \, dx = \frac{\sin(8x)}{8} \)

Putting it all together:
\[ = \frac{1}{8} \left( 3x + 4 \cdot \frac{\sin(4x)}{4} + \frac{\sin(8x)}{8} \right) \]
\[ = \frac{1}{8} \left( 3x + \sin(4x) + \frac{\sin(8x)}{8} \right) \]
\[ = \frac{3x}{8} + \frac{\sin(4x)}{8} + \frac{\sin(8x)}{64} + C \]

### Final Answer

Thus, the final result of the integral is:
\[ \int \cos^4(2x) \, dx = \frac{3x}{8} + \frac{\sin(4x)}{8} + \frac{\sin(8x)}{64} + C \]