`intcos^(-1) ((1+tan^(2)x)/(1+tan^(2) x)) dx`

Answer

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

### Step 1: Simplify the Expression Inside the Integral
We know that:
\[ \frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos(2x) \]
This is derived from the double angle formula for cosine. Therefore, we can rewrite the integral as:
\[ \int \cos^{-1}(\cos(2x)) \, dx \]

### Step 2: Apply the Property of Inverse Cosine
Using the property that \( \cos^{-1}(\cos(y)) = y \) for \( y \) in the range of \( [0, \pi] \), we can simplify:
\[ \int \cos^{-1}(\cos(2x)) \, dx = \int 2x \, dx \]

### Step 3: Integrate
Now, we can integrate \( 2x \):
\[ \int 2x \, dx = 2 \cdot \frac{x^2}{2} + C = x^2 + C \]

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

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