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Problem with Integral question

  1. Nov 16, 2008 #1
    If I know that [tex]\int^{\pi /2}_{0}x^2 cos^2 x dx = \frac{\pi^3}{48} - \frac{pi}{8}[/tex] can I use this to calculate [tex]\int^{\pi / 2}_{- \pi /2} u^2 cos^2 u du[/tex]?
  2. jcsd
  3. Nov 16, 2008 #2


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    Re: Integral

    Obviously, since the integrand is even, the latter integral is just twice the first.
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