`int tan^(-1)(cosecx - cotx)dx = ?`

Options

A. `(x^(2))/(4) + C`
B. `(-x^(2))/(4) + C`
C. `(x^(2))/(2) + C`
D. `(-x^(2))/(2) + C`

Explanation

To solve the integral \( \int \tan^{-1}(\csc x - \cot x) \, dx \), we can follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent
We start with the expression inside the inverse tangent:
\[ \tan^{-1}(\csc x - \cot x) \]
Recall that:
\[ \csc x = \frac{1}{\sin x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \]
Thus,
\[ \csc x - \cot x = \frac{1 - \cos x}{\sin x} \]

### Step 2: Use the Identity for \( \tan^{-1} \)
Using the identity for the inverse tangent:
\[ \tan^{-1}(u) = \tan^{-1}\left(\frac{a}{b}\right) \quad \text{where } u = \frac{a}{b} \]
we can rewrite our expression:
\[ \tan^{-1}\left(\frac{1 - \cos x}{\sin x}\right) \]

### Step 3: Recognize the Form
Notice that:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
and
\[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]
Thus,
\[ \tan^{-1}\left(\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) \]

### Step 4: Simplify Further
Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the principal range:
\[ \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) = \frac{x}{2} \]

### Step 5: Integrate
Now we can integrate:
\[ \int \tan^{-1}(\csc x - \cot x) \, dx = \int \frac{x}{2} \, dx \]
This integral can be computed as:
\[ = \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{4} + C \]

### Final Answer
Thus, the final result is:
\[ \int \tan^{-1}(\csc x - \cot x) \, dx = \frac{x^2}{4} + C \]

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