Discussion
Walli's formula gives \( \displaystyle\int_{0}^{\pi/2} \sin^{2}x\,dx \) as
Answer: B
Use \( \sin^{2}x = \tfrac{1}{2}(1 - \cos 2x) \), whose second term integrates to zero over the quarter period:
\[ \int_{0}^{\pi/2}\sin^{2}x\,dx \;=\; \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi/2} \;=\; \frac{1}{2}\cdot\frac{\pi}{2} \;=\; \frac{\pi}{4}. \]
By symmetry \( \cos^{2}x \) gives the same value, and the two together sum to \( \pi/2 \)
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