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(A) Let $I = \int_{-1/2}^{1/2} f(x) dx$,where $f(x) = (\cos x) \left[ \log \left( \frac{1-x}{1+x} \right) \right]$.Check if the function is odd or eve

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(A) Let $I = \int_{-1/2}^{1/2} f(x) dx$,where $f(x) = (\cos x) \left[ \log \left( \frac{1-x}{1+x} \right) \right]$.Check if the function is odd or even:$f(-x) = \cos(-x) \left[ \log \left( \frac{1-(-x)}{1+(-x)} \right) \right]$Since $\cos(-x) = \cos x$ and $\log \left( \frac{1+x}{1-x} \right) = \log \left( \left( \frac{1-x}{1+x} \right)^{-1} \right) = -\log \left( \frac{1-x}{1+x} \right)$,We have $f(-x) = \cos x \left[ -\log \left( \frac{1-x}{1+x} \right) \right] = -f(x)$.Since $f(x)$ is an odd function and the limits of integration are symmetric $[-a, a]$,the integral is $0$.

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

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