$\int_{-\pi/2}^{\pi/2} \sin^3 x\,dx$ equals:
A$0$
B$2/3$
C$\pi$
D$1$
Answer & Solution
Correct answer: A. $0$
$\sin^3(-x) = -\sin^3 x$, so $\sin^3 x$ is an **odd** function. By Property VI, the integral over the symmetric interval $[-\pi/2, \pi/2]$ is $0$.
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