`int(1+cos^(2)x)/(1-cos2x)dx=`

Options

A. `-cotx-(x)/(2)+c`
B. `cotx+(x)/(2)+c`
C. `cotx-(x)/(2)+c`
D. `-cotx+(x)/(2)+c`

Answer

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

### Step 1: Rewrite the denominator
We know that \( \cos 2x = 1 - 2\sin^2 x \). Therefore, we can rewrite the denominator:
\[ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \]

### Step 2: Substitute the denominator into the integral
Now, substituting this into the integral, we have:
\[ \int \frac{1 + \cos^2 x}{2 \sin^2 x} \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ \int \frac{1}{2 \sin^2 x} \, dx + \int \frac{\cos^2 x}{2 \sin^2 x} \, dx \]
This simplifies to:
\[ \frac{1}{2} \int \csc^2 x \, dx + \frac{1}{2} \int \cot^2 x \, dx \]

### Step 4: Integrate the first part
The integral of \( \csc^2 x \) is:
\[ \int \csc^2 x \, dx = -\cot x \]
Thus,
\[ \frac{1}{2} \int \csc^2 x \, dx = -\frac{1}{2} \cot x \]

### Step 5: Integrate the second part
The integral of \( \cot^2 x \) can be rewritten using the identity \( \cot^2 x = \csc^2 x - 1 \):
\[ \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x - x \]
Thus,
\[ \frac{1}{2} \int \cot^2 x \, dx = \frac{1}{2} (-\cot x - x) = -\frac{1}{2} \cot x - \frac{1}{2} x \]

### Step 6: Combine the results
Now, combining both parts, we have:
\[ -\frac{1}{2} \cot x - \frac{1}{2} \cot x - \frac{1}{2} x = -\cot x - \frac{1}{2} x \]

### Step 7: Add the constant of integration
Finally, we add the constant of integration \( C \):
\[ -\cot x - \frac{1}{2} x + C \]

Thus, the final answer is:
\[ \int \frac{1 + \cos^2 x}{1 - \cos 2x} \, dx = -\cot x - \frac{1}{2} x + C \]

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