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

  • Thread starter ckwan48
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  • #1
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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!
 
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  • #2
rock.freak667
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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
 
  • #3
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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!
 
  • #4
rock.freak667
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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!
Can you show your steps? It might be hard to give you a hint when I don't really know where you get lost.
 
  • #5
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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!
 

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  • #6
Office_Shredder
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It might help to do the substitution u = (x+3)/2
 
  • #7
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i figured it out! i used u-substitution to make the integral simpler and then i factored out the 4.
 

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