Prove that $\int_{0}^{\frac{\pi}{2}} \sin^{3} x \, dx = \frac{2}{3}$.

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \sin^{3} x \, dx$.
We can write $\sin^{3} x$ as $\sin^{2} x \cdot \sin x$.
Using the identity $\sin^{2} x = 1 - \cos^{2} x$,we get:
$I = \int_{0}^{\frac{\pi}{2}} (1 - \cos^{2} x) \sin x \, dx$.
Let $u = \cos x$,then $du = -\sin x \, dx$,or $\sin x \, dx = -du$.
When $x = 0$,$u = \cos(0) = 1$.
When $x = \frac{\pi}{2}$,$u = \cos(\frac{\pi}{2}) = 0$.
Substituting these into the integral:
$I = \int_{1}^{0} (1 - u^{2}) (-du) = \int_{0}^{1} (1 - u^{2}) \, du$.
Integrating with respect to $u$:
$I = [u - \frac{u^{3}}{3}]_{0}^{1} = (1 - \frac{1}{3}) - (0 - 0) = \frac{2}{3}$.
Thus,the result is proved.