The value of `int_(0) ^(2pi) cos ^(-1) ((1- tan ^(2) ""(x)/(2 ))/(1+ tan ^(2)""(x)/(2)))` dx is:

Options

A. `pi^(2)`
B. `(pi^(2))/(2)`
C. `2pi^(2)`
D. `pi^(3)`

Answer

To solve the integral

\[ I = \int_{0}^{2\pi} \cos^{-1} \left( \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \right) dx, \]

we can use the identity

\[ \cos x = \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)}. \]

### Step 1: Substitute the identity
Using the identity, we can rewrite the integral as:

\[ I = \int_{0}^{2\pi} \cos^{-1}(\cos x) \, dx. \]

### Step 2: Simplify the integral
The function \(\cos^{-1}(\cos x)\) gives us \(x\) for \(x\) in the interval \([0, \pi]\) and \(2\pi - x\) for \(x\) in the interval \([\pi, 2\pi]\). Therefore, we can break the integral into two parts:

\[ I = \int_{0}^{\pi} x \, dx + \int_{\pi}^{2\pi} (2\pi - x) \, dx. \]

### Step 3: Evaluate the first integral
Calculating the first integral:

\[ \int_{0}^{\pi} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\pi} = \frac{\pi^2}{2}. \]

### Step 4: Evaluate the second integral
Now, calculating the second integral:

\[ \int_{\pi}^{2\pi} (2\pi - x) \, dx = \int_{\pi}^{2\pi} 2\pi \, dx - \int_{\pi}^{2\pi} x \, dx. \]

Calculating the first part:

\[ \int_{\pi}^{2\pi} 2\pi \, dx = 2\pi \left[ x \right]_{\pi}^{2\pi} = 2\pi (2\pi - \pi) = 2\pi^2. \]

Calculating the second part:

\[ \int_{\pi}^{2\pi} x \, dx = \left[ \frac{x^2}{2} \right]_{\pi}^{2\pi} = \frac{(2\pi)^2}{2} - \frac{\pi^2}{2} = \frac{4\pi^2}{2} - \frac{\pi^2}{2} = \frac{3\pi^2}{2}. \]

### Step 5: Combine the results
Now we can combine the results of the two integrals:

\[ I = \frac{\pi^2}{2} + \left( 2\pi^2 - \frac{3\pi^2}{2} \right) = \frac{\pi^2}{2} + 2\pi^2 - \frac{3\pi^2}{2} = \frac{\pi^2}{2} + \frac{4\pi^2}{2} - \frac{3\pi^2}{2} = \frac{2\pi^2}{2} = \pi^2. \]

### Final Answer
Thus, the value of the integral is:

\[ I = 2\pi^2. \]