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

Answer

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

### Step 1: Simplify the integrand
We start with the expression:
\[ \frac{1 - \tan^2 x}{1 + \tan^2 x} \]
Using the identity \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \) and \( 1 + \tan^2 x = \sec^2 x \), we can rewrite the integrand:
\[ \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{1 - \tan^2 x}{\sec^2 x} \]
Since \( \sec^2 x = \frac{1}{\cos^2 x} \), we can express this as:
\[ \frac{1 - \tan^2 x}{\sec^2 x} = (1 - \tan^2 x) \cos^2 x \]

### Step 2: Rewrite the expression
Now, we can rewrite the integrand:
\[ (1 - \tan^2 x) \cos^2 x = \cos^2 x - \tan^2 x \cos^2 x \]
Since \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \), we have:
\[ \tan^2 x \cos^2 x = \sin^2 x \]
Thus, the integrand becomes:
\[ \cos^2 x - \sin^2 x \]

### Step 3: Use the double angle identity
We know that:
\[ \cos^2 x - \sin^2 x = \cos 2x \]
So we can rewrite the integral as:
\[ \int (\cos^2 x - \sin^2 x) \, dx = \int \cos 2x \, dx \]

### Step 4: Integrate
The integral of \( \cos 2x \) is:
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x + C \]

### Step 5: Final expression
Thus, we can express the final answer as:
\[ \frac{1}{2} \sin 2x + C \]