The value of `({int_(-1//2)^(1//2) cos2x.log((1+x)/(1-x))dx})/({int_(0)^(1//2)cos2x.log((1+x)/(1-x))dx})`is

Options

A. 0
B. 4
C. 2
D. 8

Explanation

To solve the given problem, we need to evaluate the expression:

\[ \frac{\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(2x) \log\left(\frac{1+x}{1-x}\right) dx}{\int_{0}^{\frac{1}{2}} \cos(2x) \log\left(\frac{1+x}{1-x}\right) dx} \]

### Step 1: Analyze the Numerator

Let \( N = \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(2x) \log\left(\frac{1+x}{1-x}\right) dx \).

### Step 2: Check if the Function is Odd or Even

We will check if the function \( f(x) = \cos(2x) \log\left(\frac{1+x}{1-x}\right) \) is odd or even.

1. Calculate \( f(-x) \):
\[ f(-x) = \cos(-2x) \log\left(\frac{1-x}{1+x}\right) = \cos(2x) \log\left(\frac{1-x}{1+x}\right) \]
Using the property of logarithms:
\[ \log\left(\frac{1-x}{1+x}\right) = \log\left(\frac{1+x}{1-x}\right)^{-1} = -\log\left(\frac{1+x}{1-x}\right) \]
Therefore,
\[ f(-x) = \cos(2x)(-\log\left(\frac{1+x}{1-x}\right)) = -f(x) \]

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

### Step 3: Evaluate the Integral Over Symmetric Limits

Using the property of integrals of odd functions:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
for any odd function \( f(x) \). Thus,
\[ N = \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx = 0 \]

### Step 4: Analyze the Denominator

Let \( D = \int_{0}^{\frac{1}{2}} \cos(2x) \log\left(\frac{1+x}{1-x}\right) dx \).

Since \( D \) is an integral of a finite value (as \( \cos(2x) \) and \( \log\left(\frac{1+x}{1-x}\right) \) are both well-defined and finite in the interval \( [0, \frac{1}{2}] \)), \( D \) will yield a non-zero finite value.

### Step 5: Evaluate the Expression

Now substituting back into the original expression:
\[ \frac{N}{D} = \frac{0}{D} = 0 \]

### Final Answer

Thus, the value of the given expression is:

\[ \boxed{0} \]