The value of the integral `int_(-1//2)^(1//2) cos x log ((1+x)/(1-x)) dx`

Options

A. 0
B. `1//2`
C. `-1//2`
D. None of these

Answer

To solve the integral

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

we will use the property of definite integrals and the symmetry of the integrand.

### Step 1: Define the function
Let

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

### Step 2: Find \( f(-x) \)
Now, we need to evaluate \( f(-x) \):

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

Using the properties of cosine and logarithm, we have:

\[ \cos(-x) = \cos x, \]
and

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

Thus,

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

### Step 3: Check the property of the integral
Since \( f(-x) = -f(x) \), we can use the property of definite integrals:

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

In our case, we have \( a = \frac{1}{2} \), so:

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

### Conclusion
Thus, the value of the integral is

\[ \boxed{0}. \]