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(B) Let $I = \int_0^\pi x \log(\sin x) \, dx$ ..... $(i)$ Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\p

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$\frac{\pi^2}{2} \log(\frac{1}{2})$

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(B) Let $I = \int_0^\pi x \log(\sin x) \, dx$ ..... $(i)$ Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\pi - x) \log(\sin(\pi - x)) \, dx = \int_0^\pi (\pi - x) \log(\sin x) \, dx$ ..... $(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^\pi (x + \pi - x) \log(\sin x) \, dx = \pi \int_0^\pi \log(\sin x) \, dx$ Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: $2I = 2\pi \int_0^{\pi/2} \log(\sin x) \, dx$ $I = \pi \int_0^{\pi/2} \log(\sin x) \, dx$ We know that $\int_0^{\pi/2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 = \frac{\pi}{2} \log(\frac{1}{2})$. Therefore,$I = \pi \left( \frac{\pi}{2} \log(\frac{1}{2}) \right) = \frac{\pi^2}{2} \log(\frac{1}{2})$.

$\int_0^\pi x \log(\sin x) \, dx = $

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