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Long integral

  1. Feb 19, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]\int \frac{t^{3}}{\sqrt{3 + t^{2}}}[/itex]

    2. Relevant equations

    ∫udv = uv - ∫vdu

    3. The attempt at a solution
    So I tried integration by parts, then I had to integrate the last term with the same method, and then I got a u substitution integral, in the end I got.

    [itex]\int \frac{t^{3}}{\sqrt{3 + t^{2}}}[/itex] = [itex]\frac{t^{4}}{\sqrt{3+t^{2}}\cdot 4} - \frac{t^{3}}{\sqrt{3+t^{2}}\cdot 2} - \sqrt[3/2]{3+t^{2}} +9\cdot \sqrt{3+t^{2}}[/itex]

    This seems a little long for the space given for the question, so could someone confirm the correctness of this integral? Also how would I go about being more efficient in solving my integrals?
     
    Last edited: Feb 19, 2014
  2. jcsd
  3. Feb 19, 2014 #2
    Are you sure this integral is copied correctly? The answer isn't pretty and is longer than your answer. In the future, try to include all of your intermediate steps as well.

    Are you sure one of the ##t^3## isn't supposed to be a ##t^2##?
     
  4. Feb 19, 2014 #3
    Don't let the space provided in a question throw you off. It has no value when solving problems. Don't get caught up in head games.

    If you rewrite the question as [itex]\int t^3 \cdot (3+t^3)^\frac{-1}{2} dt[/itex], it might be a bit easier.

    The only way I can see this problem being solved is integration by parts numerous times to eliminate the t3 term.
    It's very painful but is good practice.
     
  5. Feb 19, 2014 #4
    I corrected it, my question was wrong.
     
  6. Feb 19, 2014 #5

    SammyS

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    That makes a huge difference.

    Show how you do the first integration by parts.

    Added in Edit:

    See Dick's reply (next). That should work fine.
     
    Last edited: Feb 19, 2014
  7. Feb 19, 2014 #6

    Dick

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    Alright, then just start with the substitution u=3+t^2. I don't think you need integration by parts at all. Show your work.
     
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