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

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$\pi$

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(A) Let $I = \int_0^\pi x \sin x \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\pi - x) \sin(\pi - x) \, dx$ Since $\sin(\pi - x) = \sin x$,we have: $I = \int_0^\pi (\pi - x) \sin x \, dx = \pi \int_0^\pi \sin x \, dx - \int_0^\pi x \sin x \, dx$ $I = \pi \int_0^\pi \sin x \, dx - I$ $2I = \pi \int_0^\pi \sin x \, dx$ $2I = \pi [-\cos x]_0^\pi$ $2I = \pi [-\cos \pi - (-\cos 0)]$ $2I = \pi [-(-1) - (-1)] = \pi [1 + 1] = 2\pi$ $I = \pi$.

$\int_0^\pi x \sin x \, dx = $

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