Question

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

Hint:

To simplify definite integrals, check whether the integrand is an odd or even function. Integrals of odd functions over symmetric intervals are always zero.

Answer 1

Step 1: Let \( f(x) = \cos x \cdot \log \left( \frac{1 - x}{1 + x} \right) \). To evaluate the integral, observe the function \( f(x) \). For \( f(-x) \): \[ f(-x) = \cos(-x) \cdot \log \left( \frac{1 - (-x)}{1 + (-x)} \right) \] \[ = \cos x \cdot \log \left( \frac{1 + x}{1 - x} \right) \] \[ = -\cos x \cdot \log \left( \frac{1 - x}{1 + x} \right) \] \[ f(-x) = -f(x) \quad \text{(odd function)}. \] Step 2: Property of definite integrals for odd functions. For any odd function \( f(x) \), the integral over a symmetric interval \( [-a, a] \) is zero: \[ \int_{-a}^a f(x) dx = 0. \] Here, since \( f(x) = \cos x \cdot \log \left( \frac{1 - x}{1 + x} \right) \) is odd, we have: \[ I = \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos x \cdot \log \left( \frac{1 - x}{1 + x} \right) dx = 0. \] Conclusion: The value of the integral is: \[ I = 0. \]