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Help with integral (pretty easy but my brain is stuck!)

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data

    This is to find the arc length of r= [tex]\theta[/tex] from 0 < [tex]\theta[/tex] < 2pi

    I ended up with [tex]\int[/tex][tex]\sqrt{\theta^2 + 1}[/tex]

    2. Relevant equations

    [tex]\int[/tex][tex]\sqrt{\theta^2 + 1}[/tex]


    3. The attempt at a solution

    I couldn't see a relevant way to do integration by parts so I went with trig substitution, subbing tan(t) for theta. I got lost at the point where I'm to integrate sec^3(t)dt.

    Can anyone help? Am I on the right track?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 29, 2008 #2
    Looks good to me.

    [tex]\int\sec^{3}tdt[/tex]

    Let's start by breaking secant to the cube.

    [tex]\int\sec t\sec^2 tdt[/tex]

    Which would you make u and dV?
     
  4. Jan 29, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You actually mean [tex]\int\sqrt{\theta^2+ 1} d\theta[/tex] don't you?


    3. The attempt at a solution

    I couldn't see a relevant way to do integration by parts so I went with trig substitution, subbing tan(t) for theta. I got lost at the point where I'm to integrate sec^3(t)dt.

    Can anyone help? Am I on the right track?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data
    Looks to me like a pretty standard integral. Let [itex]\theta= tan(x)[/itex] so that [itex]d\theta= sec^2(x) dx[/itex] and [itex]\sqrt{\theta^2+ 1}= \sqrt{tan^2(x)+ 1}= sec(x)[/itex]. Then
    [tex]\int\sqrt{\theta^2+ 1}d\theta= \int sec^3(x)dx= \int \frac{1}{cos^3(x)}dx[/tex]
    [tex]= \int\frac{cox(x)}{cos^3(x)}dx= \int\frac{(cos(x)dx)}{(1- sin^2(x))^2}[/tex]

    and the substitution u= sin(x) converts to a rational integral:
    [tex]\int \frac{1}{(1-u^2)^2}du[/itex]


    2. Relevant equations



    3. The attempt at a solution[/QUOTE]
     
  5. Jan 29, 2008 #4
    ahhhh...thank you. it's relatively early in the semester and i may have partied too hard over winter break ;)
     
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