A) 0
B) \[3/8\]
C) \[8/3\]
D) \[\pi \]
Correct Answer: C
Solution :
We have \[2\int\limits_{0}^{\pi }{{{\sin }^{3}}\theta }\,d\theta =2\int\limits_{0}^{\pi }{\frac{(3\sin \theta -\sin 3\theta )}{4}}\,d\theta \] \[=\frac{1}{2}\left[ -3\cos \theta +\frac{\cos 3\theta }{3} \right]_{0}^{\pi }=\frac{1}{2}\left[ -3(-1-1)+\frac{(-1-1)}{3} \right]=\frac{8}{3}\].
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