intlimits_0^pi (x cdot sin^2 x cdot cos x) d | vedclass

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(A) Let $I = \int\limits_0^\pi  x \sin^2 x \cos x \, dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi (\pi - x

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(A) Let $I = \int\limits_0^\pi  x \sin^2 x \cos x \, dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi (\pi - x) \sin^2(\pi - x) \cos(\pi - x) \, dx$ Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get: $I = \int_0^\pi (\pi - x) \sin^2 x (-\cos x) \, dx$ $I = -\pi \int_0^\pi \sin^2 x \cos x \, dx + \int_0^\pi x \sin^2 x \cos x \, dx$ $I = -\pi \int_0^\pi \sin^2 x \cos x \, dx + I$ This implies $\pi \int_0^\pi \sin^2 x \cos x \, dx = 0$. Let $u = \sin x$,then $du = \cos x \, dx$. When $x=0, u=0$ and when $x=\pi, u=0$. Thus,$\int_0^\pi \sin^2 x \cos x \, dx = \int_0^0 u^2 \, du = 0$. Therefore,$I = 0$.

$\int\limits_0^\pi (x \cdot \sin^2 x \cdot \cos x) dx =$

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