A) \[\frac{4\pi }{3}\]
B) \[\frac{2\pi }{3}\]
C) 0
D) None of these
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi }{x{{\sin }^{3}}x\,dx}\] ?..(i) Also \[I=\int_{0}^{\pi }{(\pi -x){{\sin }^{3}}x\,\,dx}\] ?..(ii) Adding (i) and (ii), we get \[2I=\pi \int_{0}^{\pi }{{{\sin }^{3}}x}\,\,dx=\frac{\pi }{4}\int_{0}^{\pi }{\{3\sin x-\sin 3x\}dx}\] \[=\frac{\pi }{4}\left[ -3\cos x+\frac{\cos 3x}{3} \right]_{0}^{\pi }=\frac{\pi }{4}\left[ 3-\frac{1}{3}+3-\frac{1}{3} \right]=\frac{4\pi }{3}\] Hence, \[I=\frac{2\pi }{3}\].
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