`int_(-1//2) ^(1//2) (cos x ) log ((1-x)/(1+x)) dx = `

Options

A. 0
B. 1
C. `sqrt( e)`
D. `2sqrt( e)`

Answer

To solve the definite integral

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

we will follow these steps:

### Step 1: Define the function
Let

\[ f(x) = \cos x \log \left(\frac{1-x}{1+x}\right). \]

### Step 2: Check if the function is odd or even
To determine whether \( f(x) \) is odd or even, we need to evaluate \( f(-x) \):

\[ f(-x) = \cos(-x) \log \left(\frac{1+x}{1-x}\right). \]

Since \( \cos(-x) = \cos x \), we have:

\[ f(-x) = \cos x \log \left(\frac{1+x}{1-x}\right). \]

### Step 3: Simplify \( f(-x) \)
Using the properties of logarithms, we can rewrite \( f(-x) \):

\[ f(-x) = \cos x \left(-\log \left(\frac{1-x}{1+x}\right)\right) = -\cos x \log \left(\frac{1-x}{1+x}\right) = -f(x). \]

### Step 4: Conclusion about the function
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 5: Use the property of definite integrals
The integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

Thus, we can conclude:

\[ I = \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx = 0. \]

### Final Answer
Therefore, the value of the integral is

\[ \boxed{0}. \]

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