allery$\int_0^{\pi/2} \sin^2 x\,dx$ equals:
A$\pi/2$
B$\pi/4$
C$1$
D$1/2$
Answer & Solution
Correct answer: B. $\pi/4$
Use $\sin^2 x = (1 - \cos 2x)/2$.
Then $\int_0^{\pi/2} (1 - \cos 2x)/2\,dx = (1/2)[x - \sin 2x/2]_0^{\pi/2} = (1/2)[\pi/2 - 0 - 0 + 0] = \pi/4$.
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