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Calculus problem: Integral of sqrt(1/4-(x+3)^2

  1. Sep 17, 2013 #1
    I've been trying to solve this integral for a very long time!
    Problem: Integral of sqrt(1/4-(x+3)^2

    Effort: I have tried multiplying by sqrt(1/4) to get rid of the 4 so it can become a 1, so it will match up with inverse sine's derivative. After that I got kind of lost, since I don't know what happens or what will (x+3)^2, will become. I know what the answer should be like, but I want to do it without using that special formula, that they provide.

    I know there's some trick because I didn't in high school, however, I don't recall it. I don't want to use that simple formula because it doesn't really tell you what's happening in the integration process. Can someone help me out? Thanks!
    Last edited: Sep 17, 2013
  2. jcsd
  3. Sep 17, 2013 #2


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    [tex] \int \sqrt{\frac{1}{4-(x+3)^2}}dx [/tex]

    If that is your integral, you could simplify this to

    [tex] \int \frac{1}{\sqrt{4-(x+3)^2}}dx[/tex]

    Then try a trig substitution.

    Think of sin2A+cos2A=1
  4. Sep 17, 2013 #3
    Yeah that is, I know it's suppose to be inverse sine and then I try getting rid of the 4 by multiplying sqrt(1/4) and then I get lost. Extra hints would help, thanks!
  5. Sep 17, 2013 #4


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    Can you show your steps? It might be hard to give you a hint when I don't really know where you get lost.
  6. Sep 17, 2013 #5
    I've attached a file, of what I have done so far. The "question mark" is where I'm stuck. And am I doing this step, right? Thanks!

    Attached Files:

  7. Sep 17, 2013 #6


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    It might help to do the substitution u = (x+3)/2
  8. Sep 17, 2013 #7
    i figured it out! i used u-substitution to make the integral simpler and then i factored out the 4.
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