A) \[0\]
B) \[\pi \]
C) \[4/3\]
D) \[3/8\]
Correct Answer: C
Solution :
Let \[I=\int_{0}^{\pi }{|{{\sin }^{3}}\theta |d\theta }\] Since, \[\sin \theta \] is positive in the interval\[0\]to\[\pi \]. \[\therefore \] \[I=\int_{0}^{\pi }{{{\sin }^{3}}\theta \,\,d\theta }\] \[=\int_{0}^{\pi }{(1-{{\cos }^{2}}\theta )}\sin \theta \,\,d\theta \] Put, \[\cos \theta =t\Rightarrow -\sin \theta .d\theta =dt\] \[\therefore \] \[I=-\int_{1}^{-1}{(1-{{t}^{2}})}dt=-2\left[ t-\frac{{{t}^{3}}}{3} \right]_{0}^{-1}\] \[=-2\left[ -1+\frac{1}{3} \right]=\frac{4}{3}\] Alternate Solution: Let \[I=\int_{0}^{\pi }{|{{\sin }^{3}}\theta |}\,d\theta \] \[=\int_{0}^{\pi }{{{\sin }^{3}}\theta }\,d\theta \] \[=2\int_{0}^{\pi /2}{{{\sin }^{3}}\theta \,\,d\theta }\] Use Gamma function \[=2\left[ \frac{2}{3\cdot 1} \right]=\frac{4}{3}\]
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