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Definite integral with square on bottom

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}}dt[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Ok, this is an odd problem I'm working from the book. The book says the answer is [itex]\pi[/itex].

    First I tried getting rid of the fraction:

    [itex]\frac{6}{\sqrt{1-t^2}}dt[/itex] = [itex]\frac{6}{(1-t^{2})^{1/2}}[/itex] = [itex]6(1-t^{2})^{-1/2}[/itex]

    I think I may be getting this next step wrong, but next, I did this:

    = [itex]6(1^{1/2}-t^{3/2})[/itex] = [itex]6^{1/2}-6t^{3/2}[/itex]

    Which I integrated to:

    [itex]6t^{1/2} - \frac{12}{5}t^{5/2}[/itex]

    I can't seem to plug in my numbers and get [itex]\pi[/itex] though.

    Where did I mess up so far?

    Thanks for taking the time to read through my poor attempt at solving this.
     
  2. jcsd
  3. Sep 3, 2011 #2

    micromass

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    This step is wrong. I can't even see what you did there. :frown:
    Basically, you cannot get rid of the square root by splitting it over a sum, something like that is not allowed.

    Try to do a trigonometric substitution. A subsitution like t=sin(u)
     
  4. Sep 3, 2011 #3
    [itex]6(1-t^{2})^{-1/2}[/itex]

    I tried to take the -1/2 exponent and add it to the 1-t^2 which gave me this:

    = [itex]6(1^{1/2}-t^{3/2})[/itex] = [itex]6^{1/2}-6t^{3/2}[/itex]

    I figured I was doing it wrong, but my problem is that I'm not quite sure how to integrate [itex]6(1-t^{2})^{-1/2}[/itex]

    I'm not really sure how to change it into another form either.

    I'm not sure I understand how substituting in sin would help.
     
  5. Sep 3, 2011 #4

    micromass

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    It would get rid of the square root...

    What happens if you substitute in sin(u)??
     
  6. Sep 3, 2011 #5
    factor out the 6, then the integral become arcsin(t)
     
  7. Sep 3, 2011 #6
    I don't think we have covered anything on arcsin yet. I'm only in Calculus 2. If I should know that at this level, I guess one of my teachers let me down.
     
  8. Sep 3, 2011 #7
    Well I had AP Calculus last year and we covered that, but I'm in Calculus 2 this year and we haven't gotten that far yet in class but it'll come up later. Same thing with trig substitutions as someone previously stated.
     
  9. Sep 3, 2011 #8

    SammyS

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    Have you had substitution yet?

    Do as micromass suggested. Let t=sin(u) , then dt = cos(u) du .

    Yes, that's equivalent to letting u = arcsin(t), but you won't need that fact until plugging-in the limits of integration, and at that point it should pose no problem.

    BTW, 1 - sin2(θ) = cos2(θ) .
     
  10. Sep 3, 2011 #9
    Nope. We haven't had substitutions yet either. It looks like that is the next chapter we will cover if we stick to the order of the book.
     
  11. Sep 3, 2011 #10

    micromass

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    OK, so you can't use substitution??

    Well, do you have a list of standard integrals, then??
     
  12. Sep 3, 2011 #11
    After looking through the book, I found a table of integrals that has the inverse sin function. I don't seem to have the inverse ones in my notes. I don't know how I missed that. I got it figured out now. Thanks for the help.
     
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