`int_(0)^(2pi)log((a+bsecx)/(a-bsecx))dx` equals

Options

A. `(pi(a+b))/(a-b)`
B. `(pi)/(2)`
C. `(pi)/(2)(a^(2)-b^(2))`
D. 0

Answer

To evaluate the integral

\[ I = \int_0^{2\pi} \log\left(\frac{a + b \sec x}{a - b \sec x}\right) dx, \]

we will follow these steps:

### Step 1: Define the Integral
Let

\[ I = \int_0^{2\pi} \log\left(\frac{a + b \sec x}{a - b \sec x}\right) dx. \]

### Step 2: Check for Symmetry
We will check if the function is even by evaluating \(I\) at \(x\) and \(-x\):

\[ f(-x) = \log\left(\frac{a + b \sec(-x)}{a - b \sec(-x)}\right) = \log\left(\frac{a + b \sec x}{a - b \sec x}\right) = f(x). \]

Since \(f(-x) = f(x)\), the function is even.

### Step 3: Use the Property of Even Functions
For an even function over the interval \([0, 2\pi]\), we can simplify the integral:

\[ I = 2 \int_0^{\pi} \log\left(\frac{a + b \sec x}{a - b \sec x}\right) dx. \]

### Step 4: Change of Variables
Now, we will use the substitution \(x = \pi - t\):

\[ dx = -dt. \]

The limits change as follows:
- When \(x = 0\), \(t = \pi\).
- When \(x = \pi\), \(t = 0\).

Thus, we have:

\[ I = 2 \int_{\pi}^{0} \log\left(\frac{a + b \sec(\pi - t)}{a - b \sec(\pi - t)}\right)(-dt) = 2 \int_0^{\pi} \log\left(\frac{a - b \sec t}{a + b \sec t}\right) dt. \]

### Step 5: Combine the Integrals
Now we have two expressions for \(I\):

1. \(I = 2 \int_0^{\pi} \log\left(\frac{a + b \sec x}{a - b \sec x}\right) dx\)
2. \(I = 2 \int_0^{\pi} \log\left(\frac{a - b \sec x}{a + b \sec x}\right) dx\)

Adding these two equations:

\[ 2I = 2 \int_0^{\pi} \left(\log\left(\frac{a + b \sec x}{a - b \sec x}\right) + \log\left(\frac{a - b \sec x}{a + b \sec x}\right)\right) dx. \]

Using the property of logarithms, we combine the logs:

\[ \log\left(\frac{(a + b \sec x)(a - b \sec x)}{(a - b \sec x)(a + b \sec x)}\right) = \log(1) = 0. \]

### Step 6: Evaluate the Integral
Thus, we have:

\[ 2I = 2 \int_0^{\pi} 0 \, dx = 0. \]

This implies:

\[ I = 0. \]

### Final Result
The value of the integral is

\[ \int_0^{2\pi} \log\left(\frac{a + b \sec x}{a - b \sec x}\right) dx = 0. \]