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Math Help - Help integrate cosine

  1. #1
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    Help integrate cosine

    integral of (1-cos^2 x) dx

    might be easy but i need to make sure
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Re: Help

    The trick here is to rewrite cos^2x as (1+cos(2x))/2 -- which is a trig identity.

    So we have

    int(1-cos^2x) dx = int(1 - (1+cos(2x))/2) dx
    = int(1) dx - (1/2)int(1+cos(2x)) dx
    = x - (1/2)(x + (1/2)sin(2x)) + C
    = x - x/2 -(1/4)sin(2x) + C
    = x/2 - (1/2)sin(x)cos(x) + C
    = (1/2)(x - sin(x)cos(x)) + C

    The second to last step was also a trig identity. sin(2x)=2sin(x)cos(x). So (1/4)sin(2x) = (1/4)(2sin(x)cos(x))=(1/2)sin(x)cos(x)
    Then I factored the (1/2) to make the answer look a bit neater.
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  3. #3
    Junior Member AlvinCY's Avatar
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    Quote Originally Posted by smitty946 View Post
    integral of (1-cos^2 x) dx

    might be easy but i need to make sure

    We'll use a nifty identity: cos 2x = 2cos^2 x - 1 so cos^2 x = (1/2)(cos 2x + 1)


    integral of (1-cos^2 x)
    = integral of 1 - integral of cos^2 x
    = x - integral of (1/2)(cos 2x + 1)
    = x - (1/2) integral of (cos 2x + 1)
    = x - (1/2) [integral of cos 2x + integral of 1]
    = x - (1/2) [(1/2) sin 2x + x] + Constant
    = (1/2)x - 1/4 sin 2x + Constant
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