`intx((sec 2x-1))/((sec 2x+1))dx = `

Options

A. `x tan x - log sec x -x^2/2`
B. `x tan x + log sec x +x^2/2`
C. `xsec^2x+logtan x-x^2/2`
D. none

Answer

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

### Step 1: Rewrite the integral in terms of cosine
Recall that \( \sec 2x = \frac{1}{\cos 2x} \). Therefore, we can rewrite the integral as:
\[ \int \frac{\frac{1}{\cos 2x} - 1}{\frac{1}{\cos 2x} + 1} \, dx \]
This simplifies to:
\[ \int \frac{1 - \cos 2x}{1 + \cos 2x} \, dx \]

### Step 2: Simplify the fraction
To simplify \( \frac{1 - \cos 2x}{1 + \cos 2x} \), we can multiply the numerator and denominator by \( \cos 2x \):
\[ \int \frac{(1 - \cos 2x) \cos 2x}{(1 + \cos 2x) \cos 2x} \, dx = \int \frac{\cos 2x - \cos^2 2x}{\cos 2x + \cos^2 2x} \, dx \]
This gives us:
\[ \int \frac{1 - \cos 2x}{1 + \cos 2x} \, dx \]

### Step 3: Use trigonometric identities
Using the identity \( 1 - \cos 2x = 2 \sin^2 x \) and \( 1 + \cos 2x = 2 \cos^2 x \), we can rewrite the integral:
\[ \int \frac{2 \sin^2 x}{2 \cos^2 x} \, dx = \int \tan^2 x \, dx \]

### Step 4: Integrate using the identity for tangent
The integral of \( \tan^2 x \) can be expressed as:
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
This results in:
\[ \tan x - x + C \]

### Step 5: Final answer
Thus, the final answer for the integral \( \int \frac{\sec 2x - 1}{\sec 2x + 1} \, dx \) is:
\[ \tan x - x + C \]