Explanation

(c)
Let $I = \int_{0}^{\pi}{|\sin^{3}{\theta|\ d\theta}}$
Since, $\sin\theta$ is positive in interval $(0,\ \pi)$
$\therefore\ I = \int_{0}^{\pi}{\sin^{3}{\theta\ d\theta =}}\int_{0}^{\pi}{\sin{\theta \bullet \sin^{2}{\theta\ d\theta}}}$
$= \int_{0}^{\pi}{\sin{\theta(1 - \cos^{2}{\theta)\ d\theta}}}$
$= \int_{0}^{\pi}{\sin{\theta\ d\theta + \int_{0}^{\pi}{( - \sin{\theta)\cos^{2}{\theta\ d\theta}}}}}$
$= \lbrack - {\cos{\theta\rbrack}}_{0}^{\pi} + \left\lbrack \frac{\cos^{3}\theta}{3} \right\rbrack_{0}^{\pi}$
$= - \lbrack\cos{\pi - \cos{0\rbrack + \frac{1}{3}\lbrack\cos^{3}{\pi - \cos^{3}{0\rbrack}}}}$
$= - \lbrack - 1 - 1\rbrack + \frac{1}{3}\lbrack - 1 - 1\rbrack = 2 - \frac{2}{3} = \frac{4}{3}$