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How do I integrate this: 1 to 2∫ √(25+100t^2)?

  1. Sep 29, 2011 #1
    I'm trying to find the arc length and I'm able to get this far. But it's been a long time since I've done calculus so I forgot.

    1 to 2∫ √(25+100t^2)

    I tried to do u = 25+100t^2, du = 200t dt, dt = 1/200 du, but it doesn't look right.
     
  2. jcsd
  3. Sep 29, 2011 #2

    mathman

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    dt=du/(200t), where t = √[(u-25)/100]
     
  4. Sep 29, 2011 #3

    HallsofIvy

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    That's a pretty standard trig substitution: [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex]. Divide through by [itex]cos^2(\theta)[/itex], [itex]tan^2(\theta)+ 1= cos^2(\theta)[/itex].

    Therefore, let [itex]5tan(\theta)= (1/10)t[/itex] so that [itex]tan^2(\theta)= (25/100)t^2[/itex], [itex]100t^2+ 25= 25tan^2(\theta)+ 25= 25(tan^2(\theta)+ 1)= 25cos^2(\theta)[/itex] and so [itex]\sqrt{100t^2+ 25}= 5cos(theta)[/itex].

    Of course, [itex]d(tan(\theta))= sec^2(\theta)d\theta[/itex] so that [itex](1/10)dt= 5sec^2(\theta)d\theta[/itex].
     
  5. Sep 29, 2011 #4

    Char. Limit

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    I'm pretty sure that tan^2(x)+1 is not cos^2(x). I believe Halls of Ivy meant sec^2(x).
     
  6. Sep 29, 2011 #5
    Also, if you define [itex]5\tan \theta = \frac{1}{10}t[/itex], then [itex]\tan^2 \theta = \frac{1}{2500}t^2[/itex]
     
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