`int[(cot^2 2x-1)/(2cot2x)-cos8xcot4x]dx`

Answer

To solve the integral \( \int \frac{\cot^2(2x) - 1}{2 \cot(2x) - \cos(8x) \cot(4x)} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start with the expression:
\[ \frac{\cot^2(2x) - 1}{2 \cot(2x) - \cos(8x) \cot(4x)} \]
Recall that \( \cot^2(\theta) - 1 = \csc^2(\theta) - 1 = \frac{\sin^2(\theta)}{\cos^2(\theta)} - 1 = \frac{\cos^2(\theta) - \sin^2(\theta)}{\sin^2(\theta)} \).

Thus, we can rewrite:
\[ \cot^2(2x) - 1 = \frac{\cos^2(2x) - \sin^2(2x)}{\sin^2(2x)} = \frac{\cos(4x)}{\sin^2(2x)} \]

### Step 2: Substitute in the integral
Now substituting this back into the integral gives:
\[ \int \frac{\frac{\cos(4x)}{\sin^2(2x)}}{2 \cot(2x) - \cos(8x) \cot(4x)} \, dx \]

### Step 3: Simplify the denominator
Next, we rewrite \( \cot(2x) \) and \( \cot(4x) \):
\[ \cot(2x) = \frac{\cos(2x)}{\sin(2x)}, \quad \cot(4x) = \frac{\cos(4x)}{\sin(4x)} \]
So the denominator becomes:
\[ 2 \cdot \frac{\cos(2x)}{\sin(2x)} - \cos(8x) \cdot \frac{\cos(4x)}{\sin(4x)} \]

### Step 4: Combine the terms
Now we can find a common denominator for the terms in the denominator:
\[ \frac{2 \cos(2x) \sin(4x) - \cos(8x) \cos(4x) \sin(2x)}{\sin(2x) \sin(4x)} \]

### Step 5: Substitute back into the integral
Substituting this back into the integral gives:
\[ \int \frac{\cos(4x)}{\sin^2(2x)} \cdot \frac{\sin(2x) \sin(4x)}{2 \cos(2x) \sin(4x) - \cos(8x) \cos(4x) \sin(2x)} \, dx \]

### Step 6: Simplify the integral
This simplifies to:
\[ \int \frac{\cos(4x) \sin(2x)}{2 \cos(2x) \sin(4x) - \cos(8x) \cos(4x) \sin(2x)} \, dx \]

### Step 7: Use trigonometric identities
Using the identity \( 1 - \cos(2\theta) = 2\sin^2(\theta) \), we can simplify further:
\[ \int \cot(4x) (1 - \cos(8x)) \, dx \]

### Step 8: Final integration
This can be rewritten as:
\[ \int \cot(4x) \cdot 2 \sin^2(4x) \, dx \]
Now we can integrate:
\[ \int \frac{\cos(4x)}{\sin(4x)} \cdot 2 \sin^2(4x) \, dx \]
This leads us to:
\[ \int 2 \sin(4x) \cos(4x) \, dx = \int \sin(8x) \, dx \]

### Step 9: Solve the integral
The integral of \( \sin(8x) \) is:
\[ -\frac{1}{8} \cos(8x) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ -\frac{1}{8} \cos(8x) + C \]