Explanation

(c)
Let $I = \int_{0}^{\pi}{\sin^{3}{\theta d\theta}}$
$\begin{bmatrix}
\because sin\theta > 0 \\
for\ 0 < \theta < \pi
\end{bmatrix}$
$= \int_{0}^{\pi}{\sin{\theta(1 - \cos^{2}{\theta)d\theta}}}$
Put $\cos{\theta = t\ \Rightarrow \ - \sin{\theta d\theta = dt}}$
$\therefore I = \int_{- 1}^{1}{\left( 1 - t^{2} \right)dt = \left\lbrack t - \frac{t^{3}}{3} \right\rbrack_{- 1}^{1}}$
$= 2 - \frac{2}{3}$
$= \frac{4}{3}$