`inttan^(-1){sqrt((1-cos2x)/(1+cos2x))} dx = ?`

Options

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

Answer

To solve the integral

\[ \int \tan^{-1} \left( \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \right) dx, \]

we can follow these steps:

### Step 1: Simplify the expression inside the integral

Recall the double angle identities for cosine:
\[ \cos 2x = 1 - 2\sin^2 x \quad \text{and} \quad \cos 2x = 2\cos^2 x - 1. \]

Using these identities, we can express \(1 - \cos 2x\) and \(1 + \cos 2x\):
\[ 1 - \cos 2x = 2\sin^2 x, \]
\[ 1 + \cos 2x = 2\cos^2 x. \]

Substituting these into the integral gives:
\[ \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \sqrt{\frac{\sin^2 x}{\cos^2 x}} = \frac{\sin x}{\cos x} = \tan x. \]

### Step 2: Substitute back into the integral

Now, we can rewrite the integral as:
\[ \int \tan^{-1}(\tan x) \, dx. \]

### Step 3: Simplify the integral

Since \(\tan^{-1}(\tan x) = x\) for \(x\) in the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\), we have:
\[ \int \tan^{-1}(\tan x) \, dx = \int x \, dx. \]

### Step 4: Evaluate the integral

Now we can evaluate the integral:
\[ \int x \, dx = \frac{x^2}{2} + C, \]
where \(C\) is the constant of integration.

### Final Result

Thus, the final result is:
\[ \int \tan^{-1} \left( \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \right) dx = \frac{x^2}{2} + C. \]

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