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(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$.Consider the function $f(x) = \log \left(\frac{2-\

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(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$.Consider the function $f(x) = \log \left(\frac{2-\sin x}{2+\sin x}\right)$.Now,check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left(\frac{2-\sin(-x)}{2+\sin(-x)}\right) = \log \left(\frac{2+\sin x}{2-\sin x}\right)$.Since $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we have:$f(-x) = -\log \left(\frac{2-\sin x}{2+\sin x}\right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) d x = 0$.Therefore,$I = 0$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x=$

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