The value of int_{0}^{pi} x sin^{3} x , dx is | 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 get:$I = \int_{0}^

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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 get:$I = \int_{0}^{\pi} (\pi - x) \sin^{3}(\pi - x) \, dx = \int_{0}^{\pi} (\pi - x) \sin^{3} x \, dx$ ...(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 $\sin 3x = 3 \sin x - 4 \sin^{3} x$,we have $\sin^{3} x = \frac{3 \sin x - \sin 3x}{4}$.$2I = \frac{\pi}{4} \int_{0}^{\pi} (3 \sin x - \sin 3x) \, dx$$2I = \frac{\pi}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} ight]_{0}^{\pi}$$2I = \frac{\pi}{4} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) ight]$$2I = \frac{\pi}{4} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) ight] = \frac{\pi}{4} \left[ \frac{8}{3} + \frac{8}{3} ight] = \frac{\pi}{4} 	imes \frac{16}{3} = \frac{4 \pi}{3}$$I = \frac{2 \pi}{3}$

The value of $\int_{0}^{\pi} x \sin^{3} x \, dx$ is

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