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Trig integrating with absolute values:

  1. Jul 9, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_0 ^\pi \sqrt{1-\sin^2 x} dx[/tex]


    2. Relevant equations

    [tex]1 - \sin^2 x = \cos^2 x[/tex]

    3. The attempt at a solution

    I don't know how to treat this since cos changes sign half way across the integral. I know the answer should be 2 but I keep getting 0 every which way I try.
     
  2. jcsd
  3. Jul 9, 2007 #2

    bel

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    [tex]\int_0 ^\pi \sqrt{1-\sin^2 x} dx= \int_0 ^\pi |{\cos (x)}| dx[/tex]

    So you find where the [tex] 0= \cos (x) [/tex] on the interval [tex] (0,\pi) [/tex] and then integrate separately.

    i.e, [tex]\int_0 ^\pi |{\cos (x)}| dx = \int_0 ^{\frac{\pi}{2} }\cos (x) dx - \int_{\frac{\pi}{2}}^\pi \cos (x) dx [/tex]
     
    Last edited: Jul 9, 2007
  4. Jul 9, 2007 #3
    You can split the integral over the two intervals, can't you?
     
  5. Jul 9, 2007 #4

    Gib Z

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    Homework Helper

    Are you sure :)? Certainly the area is 2, but the integral gives you the Signed area. As neutrino said, splitting the integral into 2 gives the area :), as long as you take the absolute value of the negative value you will get from the integral between pi/2 and pi.
     
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