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Evaluating a cot integral

  1. Nov 28, 2005 #1
    I have the question Evaluate [tex]\int_{Pi/4}^{Pi/8}_cot^2{2x}dx[/tex]
    So integrating this should (I hope) give [tex][\frac{-1}{2}cot{2x}-x][/tex] for those limits.
    But I have never evaluated the cot integral before, I know that cotx=1/tanx. Do I substitute this identity in and work from there?
  2. jcsd
  3. Nov 28, 2005 #2


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    You could but I suspect that [itex]\frac{1}{tan^2 x}[/itex] would not be easy to integrate. How about just [itex]cot(2x)= \frac{cos 2x}{sin 2x}[/itex]?

    [tex]\int_{\pi/4}^{\pi/8}cot^2(2x)dx= \int_{\pi/4}^{\pi/8}\frac{cos^2(2x)}{sin^2(2x)}dx[/tex].

    Since those are even powers you will need to use trig indentities to reduce them. By the way is there a reason for integrating from a larger value of x to a smaller?
  4. Nov 29, 2005 #3
    Ah that was a mistake, the x values should be the other way around.

    So would [tex]\int_{\pi/8}^{\pi/4}cot^2(2x)dx= \int_{\pi/8}^{\pi/4}\frac{cos^2(2x)}{sin^2(2x)}dx[/tex] integrate to [tex]\frac{\frac{1}{2}sin^2(2x)}{\frac{-1}{2}cos^2(2x)}[/tex] ? Which I assume would simplify to [tex]-tan^2_(2x)[/tex]
    Last edited: Nov 29, 2005
  5. Nov 29, 2005 #4


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    Unfortunately not, but use the fact that [itex]\cos ^2 \alpha + \sin ^2 \alpha = 1[/itex] on:

    [tex]\int {\frac{{\cos ^2 \left( {2x} \right)}}{{\sin ^2 \left( {2x} \right)}}dx} = \int {\frac{{1 - \sin ^2 \left( {2x} \right)}}{{\sin ^2 \left( {2x} \right)}}dx} [/tex]

    Then split the integral in two.
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