allery$\int_0^{\pi/2} \cos^2 x\,dx$ equals:
A$\pi/4$
B$\pi/2$
C$1$
D$0$
Answer & Solution
Correct answer: A. $\pi/4$
By complementary symmetry, $\int_0^{\pi/2} \sin^2 x\,dx = \int_0^{\pi/2} \cos^2 x\,dx$ (apply King).
Both sum to $\int_0^{\pi/2}(\sin^2 + \cos^2)\,dx = \int_0^{\pi/2} 1\,dx = \pi/2$.
**So each $= \pi/4$.**
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