`int(cos 4x-1)/(cot x-tanx)dx` is equal to

Options

A. `-(1)/(2)cos 4x+C`
B. `-(1)/(4)cos 4x+C`
C. `-(1)/(2)sin 2x+C`
D. None of these

Explanation

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

### Step 1: Rewrite the Denominator
We know that:
\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can rewrite the denominator:
\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \]

### Step 2: Rewrite the Integral
Now substituting the rewritten denominator into the integral gives:
\[ I = \int \frac{\cos 4x - 1}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} \, dx = \int \frac{(\cos 4x - 1) \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx \]

### Step 3: Simplify the Numerator
We can use the identity \( \cos 4x = 2\cos^2 2x - 1 \) to rewrite \( \cos 4x - 1 \):
\[ \cos 4x - 1 = 2\cos^2 2x - 2 = 2(\cos^2 2x - 1) = -2\sin^2 2x \]
Thus, the integral becomes:
\[ I = \int \frac{-2\sin^2 2x \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx \]

### Step 4: Use Substitution
Let \( T = \cos 2x \). Then, differentiating gives:
\[ dT = -2\sin 2x \, dx \quad \Rightarrow \quad dx = \frac{dT}{-2\sin 2x} \]
Substituting this into the integral, we have:
\[ I = \int \frac{-2\sin^2 2x \sin x \cos x}{\cos^2 x - \sin^2 x} \cdot \frac{dT}{-2\sin 2x} \]
This simplifies to:
\[ I = \int \frac{\sin 2x \sin x \cos x}{\cos^2 x - \sin^2 x} \, dT \]

### Step 5: Further Simplification
Using the identity \( \sin 2x = 2\sin x \cos x \), we can simplify:
\[ I = \int \frac{2\sin^2 x \cos^2 x}{\cos^2 x - \sin^2 x} \, dT \]

### Step 6: Final Integration
Now we can integrate:
\[ I = \frac{1}{2} \int \left( \frac{1}{T} - T \right) dT \]
This gives:
\[ I = \frac{1}{2} \left( \ln |T| - \frac{T^2}{2} \right) + C \]
Substituting back \( T = \cos 2x \):
\[ I = \frac{1}{2} \left( \ln |\cos 2x| - \frac{\cos^2 2x}{2} \right) + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ I = \frac{1}{2} \ln |\cos 2x| - \frac{\cos^2 2x}{4} + C \]