An integral by symmetry

MediumCalculus~4m

Evaluate ∫0π/2sin⁡2x dx\displaystyle\int_0^{\pi/2}\sin^2 x\,dx. Give a decimal to four places.

Show hints (2)+
  1. Let I=∫0π/2sin⁡2I=\int_0^{\pi/2}\sin^2. Show it equals ∫0π/2cos⁡2\int_0^{\pi/2}\cos^2 via x↦π2−xx\mapsto\tfrac{\pi}{2}-x.
  2. Add them: 2I=∫0π/21 dx2I=\int_0^{\pi/2}1\,dx.

Answer

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0.7854 (± 0.001)

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