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(C) Let $I = \int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right] dx$. Using the trigonometric identities $1-\cos 2x = 2\sin^

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(C) Let $I = \int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}ight] dx$. Using the trigonometric identities $1-\cos 2x = 2\sin^2 x$ and $1+\cos 2x = 2\cos^2 x$,we get: $I = \int_{0}^{\frac{\pi}{2}} \log \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} dx = \int_{0}^{\frac{\pi}{2}} \log \left(\frac{\sin x}{\cos x}ight) dx = \int_{0}^{\frac{\pi}{2}} \log (	an x) dx$ ... $(1)$ Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$: $I = \int_{0}^{\frac{\pi}{2}} \log \left(	an \left(\frac{\pi}{2}-xight)ight) dx = \int_{0}^{\frac{\pi}{2}} \log (\cot x) dx$ ... $(2)$ Adding equations $(1)$ and $(2)$: $2I = \int_{0}^{\frac{\pi}{2}} [\log (	an x) + \log (\cot x)] dx = \int_{0}^{\frac{\pi}{2}} \log (	an x \cdot \cot x) dx$ Since $	an x \cdot \cot x = 1$,we have: $2I = \int_{0}^{\frac{\pi}{2}} \log (1) dx = \int_{0}^{\frac{\pi}{2}} 0 dx = 0$ Therefore,$I = 0$.

$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right] dx =$

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