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Integration of sine square theta from 0 to 2pi

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Pooja

  1. sin^2θ = (1-cos2θ)/2
    so, you want
    ∫[0,2π] 1/2 (1-cos2θ) dθ = 1/2 (θ - 1/2 sin2θ) [0,2π] = π

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    Steve

  2. INT sin^2 T dT over t=0 to 2PI
    use the half angle formulas sin^2(T)= (1-cos(2T))/2

    Int .5*(T-cos(2T)dt
    = .5T - .25sin2T over limits
    = .5(2PI)-.25sin(4PI)-.5*0+.25(sin(0)=PI

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    bobpursley

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