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Homework Help: How to Integrate

  1. Feb 18, 2009 #1

    Let x= [tex]\sqrt{2}[/tex] sin[tex]\vartheta[/tex]
    dx= [tex]\sqrt{2}[/tex] cos[tex]\vartheta[/tex] d[tex]\vartheta[/tex]

    from this I got


    I think inverse substitution was not the right way to solve this problem...
    any help would be greatly appreciated! Thanks!
  2. jcsd
  3. Feb 18, 2009 #2

    Tom Mattson

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    That's fine.

    That's not fine. If [itex]x=\sqrt{2}\sin(\theta)[/itex] then shouldn't there be a sine function in the denominator? And why is there a sine function in the numerator?
  4. Feb 18, 2009 #3

    Yikes! I saw (2-x^2)^(1/2) in the numerator and used a trig identity to simplify the top...
    so after you pointed this out i corrected the mistake and am left with nothing (from my point of view) i can simplify... i have root(2)*root(sin^2(O) +1)*root(2)*cos(O)dO over (root(2)*sinO)

  5. Feb 19, 2009 #4

    Tom Mattson

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    That's not what you had in the original problem. What's going on here? :confused:
  6. Feb 19, 2009 #5
    If the posted question is the one you want to solve, think about the the trig identity relating tan and sec.
  7. Feb 19, 2009 #6
    Hey thanks! Im not sure why I didn't see this. I got it! thanks again!
  8. Feb 19, 2009 #7
    Sorry, it was a mistake on my end; I didn't give a very clear question. I used sine and cosine instead of tan and sec. whoops! it got it though but thanks for trying to help me, ill have to get better at asking if i want any help lol. :tongue:

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