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

Answer

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

### Step 1: Simplify the denominator
The denominator can be rewritten using the identities for cotangent and tangent:
\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \]
Thus, we have:
\[ I = \int \frac{\cos 4x - 1}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} \, dx = \int \frac{(\cos 4x - 1) \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx \]

### Step 2: Use trigonometric identities
We can use the identity \( \cos 4x = 2\cos^2 2x - 1 \):
\[ \cos 4x - 1 = 2\cos^2 2x - 2 = 2(\cos^2 2x - 1) = -2\sin^2 2x \]
Substituting this back into the integral gives:
\[ I = \int \frac{-2\sin^2 2x \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx \]

### Step 3: Simplify the integral
Now, we can express \( \cos^2 x - \sin^2 x \) as \( \cos 2x \):
\[ I = \int \frac{-2\sin^2 2x \sin x \cos x}{\cos 2x} \, dx \]
Next, we can express \( \sin x \cos x \) as \( \frac{1}{2} \sin 2x \):
\[ I = \int \frac{-2\sin^2 2x \cdot \frac{1}{2} \sin 2x}{\cos 2x} \, dx = -\int \frac{\sin^3 2x}{\cos 2x} \, dx \]

### Step 4: Use substitution
Let \( t = \cos 2x \). Then, \( dt = -2\sin 2x \, dx \) or \( dx = \frac{dt}{-2\sin 2x} \). We also have \( \sin 2x = \sqrt{1 - t^2} \):
\[ I = -\int \frac{\sin^3 2x}{\cos 2x} \cdot \frac{dt}{-2\sin 2x} = \frac{1}{2} \int \frac{\sin^2 2x}{\cos 2x} \, dt \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{1 - \cos^2 2x}{\cos 2x} \, dt = \frac{1}{2} \int \left( \frac{1}{\cos 2x} - \cos 2x \right) dt \]

### Step 5: Integrate
Now we can integrate:
\[ I = \frac{1}{2} \left( \ln |\sec 2x| - \frac{1}{2} \sin^2 2x \right) + C \]

### Final Step: Substitute back
Substituting back \( t = \cos 2x \):
\[ I = \frac{1}{2} \left( \ln |\sec 2x| - \frac{1}{2} (1 - \cos^2 2x) \right) + C \]

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