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Trigonometric Substitution Problem - Calculus 2

  1. Feb 5, 2009 #1
    Hey there
    This is a trig substitution for my Calculus 2 class and I really have NO idea how to get started...

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

    My professor has yet to go over how to evaluate trigonometric substitutions with coefficients in front of variables.
     
  2. jcsd
  3. Feb 5, 2009 #2
    how about factoring out the 2?
     
  4. Feb 6, 2009 #3
    How would you factor out the two?

    Because [tex]\sqrt{3-2x^2}[/tex] is not the same as 2*[tex]\sqrt{\frac{3}{2}-x^2}[/tex]
     
  5. Feb 6, 2009 #4
    Yes, but [tex] \sqrt{3-2x^2} = \sqrt{2}*\sqrt{\frac{3}{2}-x^2} [/tex]. I may not have been clear in my previous post.
     
  6. Feb 6, 2009 #5
    Oh! Duh! Thank you!!!
    I'll give it a try.
     
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