int_{-1/24}^{1/24} sec x log left(frac{1-x}{1 | vedclass

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(D) Let $I = \int_{-1/24}^{1/24} \sec x \log \left(\frac{1-x}{1+x}\right) dx$. Define $f(x) = \sec x \log \left(\frac{1-x}{1+x}\right)$. We check if t

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(D) Let $I = \int_{-1/24}^{1/24} \sec x \log \left(\frac{1-x}{1+x}\right) dx$. Define $f(x) = \sec x \log \left(\frac{1-x}{1+x}\right)$. We check if the function is odd or even by evaluating $f(-x)$: $f(-x) = \sec(-x) \log \left(\frac{1-(-x)}{1+(-x)}\right) = \sec x \log \left(\frac{1+x}{1-x}\right)$. Using the property $\log(a^{-1}) = -\log a$,we have: $f(-x) = \sec x \log \left(\frac{1-x}{1+x}\right)^{-1} = -\sec x \log \left(\frac{1-x}{1+x}\right) = -f(x)$. Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function. For an odd function,the integral over a symmetric interval $[-a, a]$ is always $0$. Therefore,$\int_{-1/24}^{1/24} f(x) dx = 0$.

$\int_{-1/24}^{1/24} \sec x \log \left(\frac{1-x}{1+x}\right) dx =$

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