Discussion
The quarter-period integral of the cubed sine, \( \int_{0}^{\pi/2}\sin^{3}x\,dx \), works out at
Answer: A
Split off one factor and substitute \( u = \cos x \), so \( du = -\sin x\,dx \):
\[ \int_{0}^{\pi/2}\sin^{3}x\,dx \;=\; \int_{0}^{1}\left(1 - u^{2}\right)du \;=\; \left[u - \frac{u^{3}}{3}\right]_{0}^{1} \;=\; \frac{2}{3}. \]
Odd powers of the sine give rational values this way, while even powers bring in \( \pi \) because their average over the quarter period is not rational
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