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Math Help - Trigo Integration

  1. #1
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    Trigo Integration

    Erm, I have this integration question:

    sqrt((1-cos(x))/2) with an upper limit of 2pie and lower limit of 0.

    I can't seem to do it right.

    I tried simplifying it:

    sqrt((1-cos(x))/2)
    = ((1-cos(x))/2)^(1/2)
    = 1 - (cos^-2(x))^(1/2)
    = 1 - cos^-1(x)

    Then I intgrated 1... which becomes x, but I have no idea how to do cos^-1(x), I saw the long formula for the integration of cos^something(x) but I don't think it applies here... because the answer is 4...
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  2. #2
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    Question

    Quote Originally Posted by Dorumonsg View Post
    ...= ((1-cos(x))/2)^(1/2)
    = 1 - (cos^-2(x))^(1/2)
    Does the above really say the following?

    \sqrt{\frac{1\, -\, \cos(x)}{2}}\, =\, 1\, -\, \sqrt{\frac{1}{\cos^2(x)}}

    If so, how did you get that?
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  3. #3
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    a sqrt is something to the power of 1/2

    and

    x/something = x^-something?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Use

    sqrt((1-cos(x))/2) = sin(x/2)
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  5. #5
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    How did u get tat?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    That's your basic half angle formula from Trig

    Note when you integrate sin(x/2) from 0 to 2pi you do get 4
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  7. #7
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    Erm...

    Sin(x/2) after integrating I have -cos(x/2), I can't get 4? And anyone knows where I can get a complete list of trigo formulas? I never really learned them before?
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  8. #8
    MHF Contributor Calculus26's Avatar
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    after integrating you have -2cos(x/2) from 0 to 2 pi

    you then have -2cos(pi) - (-2cos(0)) = -2(-1) + 2 = 4


    you should have a list of trig formulas in your Calculus book--if not

    get a trig book or Google Trig Formulas -- you should find them somewhere
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  9. #9
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    Quote Originally Posted by Calculus26 View Post
    after integrating you have -2cos(x/2) from 0 to 2 pi

    you then have -2cos(pi) - (-2cos(0)) = -2(-1) + 2 = 4


    you should have a list of trig formulas in your Calculus book--if not

    get a trig book or Google Trig Formulas -- you should find them somewhere
    How did you get -2cos(x/2)?

    If I take x/2 as U.

    And differentiate it I get -2x^-3

    So after integration I have 1/(-2x^-3)(-cos(x/2))?
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  10. #10
    MHF Contributor Calculus26's Avatar
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    if u = x/2 = (1/2) x

    du = 1/2 dx

    2du = dx

    the 1/2 is a coefficient not a power--for some reason you're

    reading x/2 as 1/(x^2) not (1/2) x
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  11. #11
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    Oh okie, I get it. But just to double check if I have a number :

    X^2, is it same as 1/X^-2

    or X^-2 = 1/X^2
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  12. #12
    MHF Contributor Calculus26's Avatar
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    yes in that case you would be correct
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