int_0^pi x sin^3 x , dx = | vedclass

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(B) Let $I = \int_0^\pi x \sin^3 x \, dx$ ..... $(i)$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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$\frac{2\pi}{3}$

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(B) Let $I = \int_0^\pi x \sin^3 x \, dx$ ..... $(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \, dx = \int_0^\pi (\pi - x) \sin^3 x \, dx$ ..... $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi (x + \pi - x) \sin^3 x \, dx = \pi \int_0^\pi \sin^3 x \, dx$Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we get $\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$.$2I = \pi \int_0^\pi \frac{3 \sin x - \sin 3x}{4} \, dx = \frac{\pi}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right]_0^\pi$$2I = \frac{\pi}{4} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) \right] = \frac{\pi}{4} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) \right]$$2I = \frac{\pi}{4} \left[ \frac{8}{3} + \frac{8}{3} \right] = \frac{\pi}{4} \left( \frac{16}{3} \right) = \frac{4\pi}{3}$Therefore,$I = \frac{2\pi}{3}$.

$\int_0^\pi x \sin^3 x \, dx = $

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