Evaluate the following: `int_(-1//2)^(1//2)cos x "log" (1-x)/(1+x)dx`

Explanation

To evaluate the 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 Integral
Let
\[ I = \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos x \log\left(\frac{1-x}{1+x}\right) \, dx. \]

### Step 2: Check for Symmetry
We will check if the function \( f(x) = \cos x \log\left(\frac{1-x}{1+x}\right) \) is odd or even. To do this, we compute \( f(-x) \):

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

Using the property \( \cos(-x) = \cos x \), we have:

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

### Step 3: Simplify \( f(-x) \)
We can rewrite \( \log\left(\frac{1+x}{1-x}\right) \) using the property of logarithms:

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

Thus,

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

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

### Step 5: Apply the Property of Integrals
We know that the integral of an odd function over a symmetric interval about zero is zero:

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

In our case, since the limits are from \(-\frac{1}{2}\) to \(\frac{1}{2}\):

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

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

\[ \boxed{0}. \]