`l=int(cos^(4)x+1)/(cotx-tanx)dx` is equal to

Options

A. `-(1)/(32)[cos^(2)2x+4cos2x+10log_(e)cos2x]+c`
B. `(1)/(32)[cos^(2)2x-4cos2x+10log_(e)cos2x]+c`
C. `(1)/(32)[cos^(2)2x-4cos2x-10log_(e)cos2x]+c`
D. none of these

Answer

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

### Step 1: Rewrite the Denominator
We start by rewriting the denominator \( \cot x - \tan x \):
\[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x} \]
Thus,
\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \]

### Step 2: Substitute in the Integral
Substituting this back into the integral, we have:
\[ I = \int \frac{\cos^4 x + 1}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} \, dx = \int \frac{(\cos^4 x + 1) \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx \]

### Step 3: Simplify the Numerator
Next, we can simplify the numerator:
\[ \cos^4 x + 1 = \cos^4 x + \sin^2 x + \cos^2 x - \sin^2 x = (\cos^2 x + 1)(\cos^2 x - \sin^2 x) \]
This allows us to rewrite the integral as:
\[ I = \int \frac{(\cos^2 x + 1)(\cos^2 x - \sin^2 x) \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx \]
The \( \cos^2 x - \sin^2 x \) terms cancel:
\[ I = \int (\cos^2 x + 1) \sin x \cos x \, dx \]

### Step 4: Use a Trigonometric Identity
Using the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \), we can rewrite the integral:
\[ I = \int \left(\cos^2 x + 1\right) \frac{1}{2} \sin 2x \, dx \]

### Step 5: Split the Integral
Now we can split the integral:
\[ I = \frac{1}{2} \int \cos^2 x \sin 2x \, dx + \frac{1}{2} \int \sin 2x \, dx \]

### Step 6: Solve Each Integral
1. For the integral \( \int \sin 2x \, dx \):
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C \]

2. For the integral \( \int \cos^2 x \sin 2x \, dx \), we can use substitution:
Let \( u = \cos x \), then \( du = -\sin x \, dx \). Thus,
\[ \int \cos^2 x \sin 2x \, dx = \int u^2 (-2u) \, du = -2 \int u^3 \, du = -\frac{2}{4} u^4 + C = -\frac{1}{2} \cos^4 x + C \]

### Step 7: Combine Results
Combining both results:
\[ I = \frac{1}{2} \left(-\frac{1}{2} \cos^4 x + C\right) + \frac{1}{2} \left(-\frac{1}{2} \cos 2x + C\right) \]
This simplifies to:
\[ I = -\frac{1}{4} \cos^4 x - \frac{1}{4} \cos 2x + C \]

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