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Oopsie, issue with change of variables to evaluate definite integral

  1. Mar 12, 2009 #1
    I needed to evaluate the following integral (for constructing Chebyshev polynomials by an orthogonalisation process), but I just discovered that I'm having an issue with the change of variable technique:P The specific integral itself is unimportant as to the issue I'm having, but by means of an example this is the one I was solving when it arose:

    [tex]
    \int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}} dx
    [/tex]

    And I believe this is done with the following change of variable:

    [tex]
    x = g(\theta) = cos (\theta); \ \ \rightarrow \ \ dx= g'(\theta) = -sin(\theta)d\theta
    [/tex]

    But since the transformation has to be bijective (or however you say:P), one would have to limit the range of [tex]\theta[/tex] otherwise the inverse relationship [tex]\theta = g^{-1}(x) [/tex] would not be a single valued function.

    So for example, I can restrict [tex]\theta[/tex] as so: [tex]
    0 \leq\theta \leq \pi
    [/tex]
    in which case [tex]g^{-1}(x=-1)=\pi ; \ \ \ g^{-1}(x=1)=0 [/tex]
    and thus:
    [tex]
    \int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}} dx = \int_{\pi}^{0} \frac{-sin(\theta)}{\sqrt{(sin(\theta)^2}} d\theta = \int_{\pi}^{0} -d\theta = \int_{0}^{\pi} d\theta = \pi
    [/tex]

    BUT! if for example I use the same transformation but restricting [tex]\theta[/tex] to: [tex]
    \pi \leq\theta \leq 2\pi
    [/tex]
    then I have: [tex]g^{-1}(x=-1)=\pi ; \ \ \ g^{-1}(x=1)=2\pi [/tex]
    and:
    [tex]
    \int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}} dx = \int_{\pi}^{2\pi} \frac{-sin(\theta)}{\sqrt{(sin(\theta)^2}} d\theta = \int_{\pi}^{2\pi} -d\theta = -\pi
    [/tex]

    I know the answer has to be [tex]\pi[/tex] because the integrand in terms of the original variable is always positive for [tex]x\in (-1,1)[/tex], thus the integral has to be positive. And I know that different signs I'm getting is because in one case my transformation is such that [tex]\theta[/tex] is increasing with increasing x (ie. [tex]g'>0[/tex]) and in the other [tex]\theta[/tex] is decreasing with increasing x (ie. [tex]g'<0[/tex]), but . . . the sign of [tex]g'<0[/tex] should compensate the change in the order of the limits of integration and I shouldn't have to worry about that, no? I just can't see what I'm doing wrong in either case, although I'm certainly doing something wrong!:(
     
  2. jcsd
  3. Mar 12, 2009 #2
    I'm not sure if this helps but I recognize that:
    [tex]\frac{d}{dx}(arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}[/tex]
    [tex]where -\frac{\pi}{2} < x < \frac{\pi}{2}[/tex]
     
  4. Mar 13, 2009 #3
    People always forget that:

    [tex] \sqrt{x^2}=|x|[/tex] and not simply x.

    Frankly, i didn't bother to look your whole work, but i think that your problem is, as i just stated:


    [tex]\sqrt{sin^2(\theta)}=|sin(\theta)|[/tex]

    and not as you, somewhere along the lines, have done simply sin(@) (@=theta)


    Now, i assume you know the definition of the abs value, right? Well, in case you cannot recall while reading this post:

    |x|=x if x>0 and -x if x<0.

    Now:

    |sin(@)|=sin(@) if sin(@)<0 and -sin(@) if sin(@)>0

    Look at your limits of integration, pi to 2pi. what do you notice???? THis is your problem..
     
  5. Mar 14, 2009 #4
    sutupidmath, . . . yes!
    that includes me :blushing:

    My problem was not with the change of variable theorem after all :P Now everything falls into place :)

    Fanku, fanku, I am much indebted :P
     
  6. Mar 14, 2009 #5
    Thanku to Chaos2009 as well for their observation:P
     
  7. Mar 14, 2009 #6
    this should have read |sin(@)|=sin@ if sin@>0 and -sin@ if sin@<0
    My bad..
     
  8. Mar 14, 2009 #7
    Lol that's okay, actually, I got your meaning without even realising you said it the wrong way round:P
     
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