`int(cos4x-1)/(cotx-tanx)dx` is equal to

Options

A. `(1)/(2)In|sec2x|-(1)/(4)cos^(2)2x+c`
B. `(1)/(2)In|sec2x|+(1)/(4)cos^(2)x+c`
C. `(1)/(2)In|cos2x|-(1)/(4)cos^(2)2x+c`
D. `(1)/(2)In|cos2x|+(1)/(4)cos^(2)x+c`

Explanation

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

### Step 1: Simplify the numerator
We know that \( \cos 4x \) can be expressed using the double angle formula:
\[ \cos 4x = 2\cos^2 2x - 1 \]
Thus, we can rewrite \( \cos 4x - 1 \):
\[ \cos 4x - 1 = 2\cos^2 2x - 1 - 1 = 2\cos^2 2x - 2 = 2(\cos^2 2x - 1) = 2(-\sin^2 2x) = -2\sin^2 2x \]

### Step 2: Simplify the denominator
The denominator \( \cot x - \tan x \) can be rewritten using sine and cosine:
\[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x} \]
Thus,
\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \]
Using the identity \( \cos^2 x - \sin^2 x = \cos 2x \), we can write:
\[ \cot x - \tan x = \frac{\cos 2x}{\sin x \cos x} \]

### Step 3: Substitute into the integral
Now substituting back into the integral:
\[ \int \frac{-2\sin^2 2x}{\frac{\cos 2x}{\sin x \cos x}} \, dx = -2 \int \frac{\sin^2 2x \cdot \sin x \cos x}{\cos 2x} \, dx \]

### Step 4: Use trigonometric identities
Using the identity \( \sin^2 2x = 1 - \cos^2 2x \):
\[ -2 \int \frac{(1 - \cos^2 2x) \sin x \cos x}{\cos 2x} \, dx \]
This can be separated into two integrals:
\[ -2 \left( \int \frac{\sin x \cos x}{\cos 2x} \, dx - \int \frac{\sin x \cos^3 2x}{\cos 2x} \, dx \right) \]

### Step 5: Use substitution
Let \( t = \cos 2x \), then \( dt = -2\sin 2x \, dx \) or \( dx = \frac{dt}{-2\sin 2x} \). Thus:
\[ \sin 2x = 2\sin x \cos x \]
Substituting gives:
\[ -2 \int \frac{\sin x \cos x}{t} \cdot \frac{dt}{-2(2\sin x \cos x)} = \int \frac{dt}{t} \]

### Step 6: Solve the integral
The integral \( \int \frac{dt}{t} \) is:
\[ \log |t| + C = \log |\cos 2x| + C \]

### Final Result
Thus, the final answer is:
\[ \int \frac{\cos 4x - 1}{\cot x - \tan x} \, dx = -\frac{1}{2} \log |\cos 2x| + C \]