A) 0
B) 3/8
C) 4/3
D) \[\pi \]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi }{|{{\sin }^{3}}\theta |}\,d\theta \] Since \[\sin \theta \] is positive in interval \[(0,\pi )\] \[\therefore I=\int_{0}^{\pi }{{{\sin }^{3}}\theta \,d\theta =\int_{0}^{\pi }{\sin \theta (1-{{\cos }^{2}}\theta )\,\,d\theta }}\] \[=\int_{0}^{\pi }{\sin \theta \,d\theta +\int_{0}^{\pi }{(-\sin \theta )\,{{\cos }^{2}}\theta \,d\theta }}\] \[=[-\cos \theta ]_{0}^{\pi }+\left( \frac{{{\cos }^{3}}\theta }{3} \right)_{0}^{\pi }=\frac{4}{3}\].
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