int_{ - 1/2}^{1/2} {cos x ln left( frac{1 + x | vedclass

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(A) Let $I = \int_{ - 1/2}^{1/2} {\cos x \ln \left( \frac{1 + x}{1 - x} \right) dx}$.Define $f(x) = \cos x \ln \left( \frac{1 + x}{1 - x} \right)$.Che

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(A) Let $I = \int_{ - 1/2}^{1/2} {\cos x \ln \left( \frac{1 + x}{1 - x} \right) dx}$.Define $f(x) = \cos x \ln \left( \frac{1 + x}{1 - x} \right)$.Check if $f(x)$ is an odd function by evaluating $f(-x)$:$f(-x) = \cos(-x) \ln \left( \frac{1 + (-x)}{1 - (-x)} \right) = \cos x \ln \left( \frac{1 - x}{1 + x} \right)$.Since $\ln \left( \frac{1 - x}{1 + x} \right) = \ln \left( \frac{1 + x}{1 - x} \right)^{-1} = -\ln \left( \frac{1 + x}{1 - x} \right)$,we have $f(-x) = -\cos x \ln \left( \frac{1 + x}{1 - x} \right) = -f(x)$.Thus,$f(x)$ is an odd function.By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$.Therefore,$I = 0$.

$\int_{ - 1/2}^{1/2} {\cos x \ln \left( \frac{1 + x}{1 - x} \right) dx}$ is equal to

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