Question

Evaluate the integral \( \int_{-\frac{1}{24}}^{\frac{1}{24}} \sec(x) \log\left(\frac{1-x}{1+x}\right) \, dx \):

• A. \(\frac{\pi}{2}\)
• B. \(\pi\)
• C. 1
• D. 0

Hint:

For integrals involving symmetric limits, always check the parity of the function (even or odd) to simplify the calculation.

Answer 1

The Correct Option is D

The function \(\log\left(\frac{1-x}{1+x}\right)\) is an odd function, and \(\sec(x)\) is an even function. When an even function is multiplied by an odd function, the result is an odd function. The integral of an odd function over a symmetric interval about zero is zero: \[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(x) \text{ is odd}. \] Here, \(f(x) = \sec(x) \log\left(\frac{1-x}{1+x}\right)\) is odd because \(\sec(x)\) is even and \(\log\left(\frac{1-x}{1+x}\right)\) is odd, which results in an odd product. Therefore: \[ \int_{-\frac{1}{24}}^{\frac{1}{24}} \sec(x) \log\left(\frac{1-x}{1+x}\right) \, dx = 0. \]