Answer
$\frac{2-\sqrt{2}}{4}$
Work Step by Step
$\int^{\frac{\pi}{8}}_{0}sin2x dx$
=$[-\frac{1}{2}cos2x]^{(\frac{\pi}{8})}_0$
=$(\frac{-1}{2}cos2((\frac{\pi}{8})))-(\frac{-1}{2}cos2(0))$
=$\frac{2-\sqrt{2}}{4}$
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