`int cot^(-1)((sin2x)/(1-cos2x)) dx = ?`

Options

A. `(-1)/((1+x^(2))) + C`
B. `(-1)/((1-x^(2))) + C`
C. `(x^(2))/(2) + C`
D. `2x^(2) + C`

Answer

To solve the integral \( \int \cot^{-1}\left(\frac{\sin 2x}{1 - \cos 2x}\right) dx \), we can follow these steps:

### Step 1: Simplify the argument of the inverse cotangent
We start with the expression inside the inverse cotangent:
\[ \frac{\sin 2x}{1 - \cos 2x} \]
Using the double angle identities, we know:
\[ \sin 2x = 2 \sin x \cos x \quad \text{and} \quad 1 - \cos 2x = 2 \sin^2 x \]
Thus, we can rewrite the expression:
\[ \frac{\sin 2x}{1 - \cos 2x} = \frac{2 \sin x \cos x}{2 \sin^2 x} = \frac{\cos x}{\sin x} = \cot x \]

### Step 2: Rewrite the integral
Now substituting back into the integral, we have:
\[ \int \cot^{-1}(\cot x) \, dx \]
Since \( \cot^{-1}(\cot x) = x \), we can simplify our integral:
\[ \int x \, dx \]

### Step 3: Integrate
The integral of \( x \) is:
\[ \frac{x^2}{2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \cot^{-1}\left(\frac{\sin 2x}{1 - \cos 2x}\right) dx = \frac{x^2}{2} + C \]