alleryIf $\displaystyle I_n = \int_0^{\pi/2} \sin^n x\, dx$, then $I_5 = ?$
A$8/15$
B$2/3$
C$3\pi/16$
D$1$
Answer & Solution
Correct answer: A. $8/15$
**Reduction (Wallis):** $I_n = \tfrac{n-1}{n} I_{n-2}$.
$I_1 = 1$, $I_3 = (2/3)(1) = 2/3$, $I_5 = (4/5)(2/3) = 8/15$.
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