The integral from pi/8 to pi/4 of (sin2x)^2
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Originally Posted by abcocoa The integral from pi/8 to pi/4 of (sin2x)^2 form basic trig identities: (sin(2x))^2 = (1/2)[1-cos(4x)] so: Integral (sin(2x))^2 dx = (1/2)x - (1/8)sin(4x) + C Integral(x=pi/8, pi/4) (sin(2x))^2 dx = [(1/2)pi/4] - [(1/8)pi/8 - (1/8)sin(pi/2)=pi/16+1/8 RonL
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