MCQ
$\int_{\,0}^{\,\pi } {\,\left| {\,{{\sin }^3}\theta \,} \right|\,d\theta } =$
- A$0$
- B$3/8$
- ✓$4/3$
- D$\pi $
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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