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The Substitution Method

  1. May 13, 2008 #1
    Evaluate the following integral using integration by substitution: [​IMG]

    Here is my attempt:
    Let x = sinu, then dx/du = cosu
    Substituting gives, ∫1/(1-sin2u)×cosu du
    = ∫1/(1-sin2u)×cosu du
    = ∫cosu/√cos2u du
    = ∫cosu/cosu du
    = ∫1 du = u + c = sin-1x + c

    Am I right? Did I get the right solution?


  2. jcsd
  3. May 13, 2008 #2
    You don't need Trig sub. for this.


    But we'll go with it!

    [tex]x=\sin u[/tex]
    [tex]dx=\cos udu[/tex]

    [tex]\int\frac{\sin u \cos u du}{\sqrt{1-\sin^2 u}}[/tex]
  4. May 13, 2008 #3
    Try u=1-x^2 instead... Might be a bit easier.
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