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Math Help - integral of trig function

  1. #1
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    integral of trig function

    Hey All,

    I wanted to know when you integrate sin^4x where the 4 comes from in the integral.

    ∫sin^4x dx= ∫(sin^2x)^2 dx= ∫(1-cos2x)^2/4 dx

    thank you
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by dadon View Post
    Hey All,

    I wanted to know when you integrate sin^4x where the 4 comes from in the integral.

    ∫sin^4x dx= ∫(sin^2x)^2 dx= ∫(1-cos2x)^2/4 dx

    thank you
    I'm sure you've seen the trig identities enough to be familiar with the double angle formulas:
    cos(2x) = 2cos^2(x) - 1 = 1 - 2sin^2(x) = cos^2(x) - sin^2(x)

    Well, for this problem, we are especially interested in the version containing only sin^2(x)
    cos(2x) = 1 - 2sin^2(x)

    Let's solve this for sin^2(x):
    sin^2(x) = 1/2(1 - cos(2x))

    Using this we can replace a sin^2(x) term (which we cannot integrate) with 1/2(1 - cos(2x)) (which we can integrate).
    Thus the integration becomes:
    ∫sin^4x dx= ∫(sin^2x)^2 dx= ∫[1/2(1 - cos(2x))]^2 dx = ∫(1/2)^2*(1 - cos(2x)^2 dx = ∫(1-cos2x)^2/4 dx
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