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Integration Involving Square Root

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int \sqrt{t^8 + t^6 } dt[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure what to do next, can someone point me in the right direction? Thank you.
     
  2. jcsd
  3. Sep 20, 2008 #2

    rock.freak667

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    write t8+t6 as t6(t2+1) and then try a trig substitution.
     
  4. Sep 20, 2008 #3

    tiny-tim

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    Hi cse63146! :smile:

    First thing would be to rewrite as:

    [tex]\int t^3\,\sqrt{1 + t^2} dt[/tex] :wink:

    EDIT: waaa! rock.freak667 beat me to it! :cry:
     
  5. Sep 20, 2008 #4
    where [tex]t = tan\vartheta \ and \ dt =sec^2 \vartheta \ d\vartheta [/tex]?
     
  6. Sep 20, 2008 #5

    Defennder

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    Yes, looks ok as an approach.
     
  7. Sep 20, 2008 #6
    stuck again (that was fast)

    so I got it to this: [tex]\int tan^6 \vartheta (sec^2 \vartheta )[/tex]

    do I use integration by parts here?

    or would I use a second substitution with u = tan and du = sec^2 ?
     
    Last edited: Sep 20, 2008
  8. Sep 20, 2008 #7

    Defennder

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    I was thinking about this a bit more and I realise you don't have to do trigo substitution. You have [tex]\int t^3 \sqrt{t^2+1}dt [/tex]. Do it by parts here, multiple times.
     
  9. Sep 21, 2008 #8

    tiny-tim

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    Hi cse63146! :smile:
    :rofl:
    Nooo … it's [tex]\int tan^3 \vartheta (sec^2 \vartheta )[/tex] …
    Easiest is probably to go to u straight from t (without going through θ), and then use integration by parts. :wink:
     
  10. Sep 21, 2008 #9
    so integration by parts is the following:

    [tex]\int f(x)'g(x) \ dx = f(x)g(x) - \int f(x)g'(x) \ dx[/tex]

    I chose f'(x) to be [tex]\sqrt{1 + t^2}[/tex] so I could eventually get rid of t^3, but I have no idea how to find f(x)

    so I need to do a trig sub like this:http://www.freemathhelp.com/forum/viewtopic.php?f=3&t=30001&start=0 but what happens to the square root? It just disappears after he uses the trig identity sin^2 + cos^2 = 1, and it's 25 instead of 5.
     
    Last edited: Sep 21, 2008
  11. Sep 22, 2008 #10

    tiny-tim

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    No … always mix-and-match to get an easy f'(x) …

    in this case choose f'(x) to be [tex]t\,\sqrt{1 + t^2}[/tex] :wink:
    No … as Defennder says, you don't need a trig sub!

    Either do integration by parts, staying with t, or (possibly slightly easier) just substitute v = 1 + t2, dv = … ? :smile:
     
  12. Sep 22, 2008 #11

    Gib Z

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    The easiest method here would probably be the appropriate hyperbolic trig substitution.
     
  13. Sep 26, 2008 #12
    dv = 2t

    so would it look like this:

    [tex]2\int t^2 \sqrt{v} dv[/tex]

    would I write dv or dt in this case since I have both t and v.

    P.S sorry it took so long to respond, I got sick shortly after this and my fever was gone this morning.
     
  14. Sep 26, 2008 #13

    tiny-tim

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    Hi cse63146! I hope you feel completely better soon. :smile:

    [tex]2\int t^2 \sqrt{v} dv[/tex] isn't quite finished, is it?

    you can make it [tex]2\int (v - 1) \sqrt{v} dv[/tex] or just [tex]2\int (v^{3/2}\,-\,v^{1/2}) dv[/tex] :wink:

    (when you make a substitution, you must change everything! :smile:)
     
  15. Sep 26, 2008 #14
    Ah, that makes sense. Once I do the integration, I just "return" t back into the equation. Thanks.
     
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