Calculus

posted by Rostrastics

Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.

  1. Steve

    Let V = ∫sin^2x dx
    Since cos^2x is a mirror image of sin^2x, ∫cos^2x dx = V

    Now, since sin^2x+cos^2x = 1,

    2V = ∫[0,π/2] 1 dx = π/2
    V = π/4

    check:

    sin^2x = (1-cos2x)/2
    ∫sin^2x dx = ∫(1-cos2x)/2 dx
    = 1/2 (x - 1/2 sin2x) [0,π/2]
    = 1/2[(π/2)-(0)]
    = π/4

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