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Trig Sub, Need check

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{\sqrt{9-x^2}}{x} dx [/tex]

    So:
    [tex]x = 3Sin(\Theta)[/tex]
    [tex]dx = 3Cos(\Theta)d\Theta[/tex]

    Rewritten:
    [tex]\int \frac{\sqrt{9 - (3Sin(\Theta))^2}*3Cos(\Theta)d\Theta}{3Sin(\Theta)}[/tex]

    [tex]\int \frac{\sqrt{9(1-Sin(\Theta)^2)}*3Cos(\Theta)d\Theta}{3Sin(\Theta)}[/tex]

    [tex]\int \frac{3Cos(\Theta)^2}{Sin(\Theta)}[/tex]

    When I check this with mathematica (replacing sin and cos with something in terms of x) it doesn't match. Its a different answer.
     
  2. jcsd
  3. Oct 6, 2008 #2
    Nevermind, I will fix that latex in a few minutes, physics forums is not working well.
     
  4. Oct 6, 2008 #3
    Ok there we go... Yes, when I integrate the initial equation in mathematica I get an answer that does not equal when I integrate at the last step I have done. And I rewrite it back in terms of x dx so that is not it. Did I do something wrong?
     
  5. Oct 6, 2008 #4

    Redbelly98

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    Your derivation looks right to me.
     
  6. Oct 6, 2008 #5
    Hmm, according to mathematica I am off exactly by a factor of 3?
     
  7. Oct 6, 2008 #6

    Dick

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    Can you say what you got and what mathematica got?
     
  8. Oct 6, 2008 #7
    Integrating at the very beginning equation mathematica gave me:

    [tex] \sqrt{9 - x^2} + 3\ln{x} - 3 \ln{3 + \sqrt{9 - x^2}}[/tex]

    and when I integrate my step I get:


    [tex] 3*( \sqrt{9 - x^2} + 3\ln{x} - 3 \ln{3 + \sqrt{9 - x^2}} )[/tex]
     
  9. Oct 6, 2008 #8
    I am also stuck at my last step, what would be a good way of integrating that?

    -but I don't know if thats right seeing how my answer doesn't match with mathematica.
     
  10. Oct 6, 2008 #9

    Redbelly98

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    How about an approximate numerical check, which will tell you who's off by a factor of 3?

    Your integrand varies wildly near x=0 and x=3, so stay away from those values. Try approximating the integral (over x) from 0.99 to 1.01, just treating the integrand as a constant evaluated at x=1. What does that give for the integral? Compare it to your final answer, and to the mathematica answer.
     
  11. Oct 6, 2008 #10
    Oh crap Im sorry, I plugged it in wrong. The answer is still different though and still wrong.

    Here is what mathematica gave me:

    [tex]-\frac{x^2}{2} + 9\ln{x}[/tex]
     
  12. Oct 6, 2008 #11
    I just tried that... plugged in 0.99 and 1.01 as my limits and got 0.0565705 from the mathematica answer and 0.160006 from the step I got to
     
  13. Oct 6, 2008 #12

    Dick

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    Your first step should be to differentiate those and see if you get sqrt(9-x^2)/x. I don't think you will, neither one looks right. Did you leave out some parentheses or something?
     
  14. Oct 6, 2008 #13
    Sorry, but that second equation changed as I realized I plugged in the wrong stuff. Still though they are different and unequal. I don't know why mathematica would be wrong and I can't figure out where I am either.
     
  15. Oct 6, 2008 #14

    Dick

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    What ARE you doing? Are you trying to turn cos(theta)^2*dtheta/sin(theta) BACK into x before you integrate? Don't do that! If you do it right, you'll just get back where you started. Integrate dtheta, THEN turn back into x.
     
  16. Oct 6, 2008 #15
    oh crap... no wonder

    thanks. I will go work on it
     
  17. Oct 6, 2008 #16

    Dick

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    Whew. Thought I was going crazy. I'll give you a free hint. cos(theta)^2=1-sin(theta)^2.
     
  18. Oct 6, 2008 #17

    Redbelly98

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    Okay. And what about the approximate answer using the original integral? I.e., what is

    [tex]
    \int \frac{\sqrt{9-1^2}}{1} dx
    [/tex]
    evaluated from 0.99 to 1.01?
     
  19. Oct 6, 2008 #18

    Dick

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    I think we've sorted it out. Somebody (and I won't name names) is new to trig substitutions. Or has forgotten.
     
  20. Oct 6, 2008 #19
    But how would you integrate the last step on my first post?
     
  21. Oct 6, 2008 #20

    Dick

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    Use cos(theta)^2=1-sin(theta)^2. I think I already suggested that. Though I think I edited it in.
     
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