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Anomalous integral

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to calculate [tex]\int_0^{2\pi} \sqrt{1-\sin{x}}\, dx[/tex] when I compute the indefinite integral I get [tex]2\sqrt{\sin{x}+1}+C[/tex] but when I evaluate it into the integration limits it turns out to be 0 wich doesn't make sence since the original function is clearly positive...I already tested this result in wolfram alpha and my calculation of the definite integral is perfect but when I tested the definite integral wolfram computed its value to be [tex]4\sqrt{2}[/tex]. I can't find out what's wrong with my procedure. Please help!


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2
    The indefinite integral you computed is not right I guess. Wolfram Alpha shows that but when I do it in Derive I get another thing which verifies the result 4sqrt2...

    This is what I get (no idea what floor is, guess it's the function?)

    $$\frac{{2\cos x}}{{\sqrt {1 - \sin x} }} - 4\sqrt 2 \cdot FLOOR\left( {\frac{1}{4} - \frac{x}{{2\pi }}} \right)$$
     
    Last edited: Apr 15, 2012
  4. Apr 15, 2012 #3
    @Hernaner28
    Floor's not even a continuous function, much less differentiable that cant possibly be a primitive for [itex] \sqrt{1-\sin{x}} [/itex].
     
  5. Apr 15, 2012 #4

    Dick

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    When you computed the antiderivative you replaced [itex]\sqrt{\cos^2 x}[/itex] with [itex]\cos x[/itex]. That's only correct if cos(x) is positive. cos(x) isn't positive on [0,2pi]. I'd suggest you split the integral up.
     
  6. Apr 16, 2012 #5
    Thank you very much guys, your contributions were very valuable; particularly thank you Dick, with your observation I solved my trouble
     
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