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A Simple Definite Integral

  1. Apr 2, 2009 #1
    I have a problem which involves finding the moment of inertia which involves evaluating: [tex]\int_0^1 x^2 \left(\frac{\pi}{2} - \sin^{-1} x\right)dx[/tex]

    Now eventually, we end up having to evaluate the integral through integration by parts: [tex]\int_0^1 \frac{x^3}{\sqrt{1-x^2}} \ dx[/tex]

    Now this is really an improper integral with the problem at [tex] x= 1[/tex] so we really have: [tex]\lim_{t \to 1^-} \int_0^t \frac{x^3}{\sqrt{1-x^2}} dx[/tex]

    Now the problem that has occurred to me is that I'm getting two different results through two different methods.

    Method 1: A u-substitution

    Let: [tex]u = \sqrt{1-x^2} \ \Rightarrow \ - u du = x dx[/tex]

    [tex]x = 0 \ \Rightarrow \ u = 1[/tex], and [tex]x \to 1 \ \Rightarrow \ u \to 0[/tex]

    So we have the new integral: [tex]\lim_{s \to 0^+} \int_s^1 \frac{1-u^2}{u} \ du = \lim_{s \to 0^+} \left(\ln u - \frac{u^2}{2}\right) \Bigg|_s^1[/tex]

    which diverges.

    Method 2: A trig-sub

    Let: [tex]x = \sin \theta \ \Rightarrow \ dx = \cos \theta d \theta[/tex]

    So, the integral becomes: [tex]\lim_{s \to \frac{\pi}{2}} \int_0^s \sin^3 \theta d \theta[/tex]

    but this is no longer improper and we do get a result.

    So what's going on here? If we do the trig-sub, we get the correct answer to the original inertia question but if we look at the function [tex]f(x) = \frac{x^3}{\sqrt{1-x^2}}[/tex], it should diverge!
  2. jcsd
  3. Apr 2, 2009 #2


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    It's hard to know without seeing your arithmetic, but I think you dropped the factor in red.
  4. Apr 2, 2009 #3
    Oh whoops! Wow what a silly mistake. Thanks!
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