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Converting degree into 'x' (integration)

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data
    After integrating x²/(sqrt1-x²) using the triangle method and thereby substituting sin¤ for x, I ended up with the result 1/2¤-1/2sin2¤+c. For my answer toge complete, I must have the ¤ in x format. I managed to convert the 1/2¤, which is simply 1/2arcsinx but I couldn't for 1/2sin2¤, my answer did not agree with the textbook's which read 1/2x(sqrt1-x²)....

    My computer won't read latex at the moment so please excuse the amateur notation. Thank you.

    2. Relevant equations
    I think I covered it all above!
     
  2. jcsd
  3. Nov 26, 2007 #2

    Dick

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    If I call your substitution sin(t) (I don't know what symbol you were trying to put in), sin(2t)=2*cos(t)*sin(t). Put t=arcsin(x) in there.
     
  4. Nov 26, 2007 #3
    Oh I see, you used the trig identity....and then I refer back to the triangle to figure out what's cos(t)...Thank you!
     
  5. Nov 26, 2007 #4
    Following your reasoning, 1/2*sin(2t) becomes 2x*sqrt(1-x²). The two cancels out when I plug it back into 1/2*sin(2t), and this can't be since the answer is 1/2*x*sqrt(1-x²)??!
     
  6. Nov 26, 2007 #5

    Dick

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    I get that the integral of sin(t)^2 is t/2-sin(2t)/4. You are missing a factor of 2.
     
  7. Nov 26, 2007 #6
    So you're suggesting that the integral of 1/2cos(2t) is 1/2sin(2t) ?! (can you please elaborate as to how you came up with the answer?)
     
  8. Nov 26, 2007 #7

    Dick

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    No, I'm suggesting it's sin(2t)/4. Substitute u=2t, du=2dt. That's your missing 2.
     
  9. Nov 26, 2007 #8
    hmm I see, I never did that before, I simply went on with the integrating whenever it was 1/2-1/2cos2t for example....and I always obtained the right answer, weird! But thank you!
     
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