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Let $I=\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$

$\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cdot \sin x d x \\ &=\int_{0}^{\frac{\pi}{2}}\left(1-\cos ^{2} x\right) \sin x d x \\ &=\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \cdot \sin x d x \\ &=[-\cos x]_{0}^{\frac{\pi}{2}}+\left[\frac{\cos ^{3} x}{3}\right]_{0}^{\frac{\pi}{2}} \\ &=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3} \end{aligned}$

Hence, the given result is proved.