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Trigonometric Substitution Proof

  1. Jul 22, 2008 #1

    LHC

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    The question is:

    Use [tex]x = \tan \theta , \frac{-\pi}{2} < \theta < \frac{\pi}{2} [/tex] to show that:

    [tex] \int_{0}^{1} \frac{x^3}{\sqrt{x^2+1}} dx =\int_{0}^{\frac{\pi}{4}} \tan^3 \theta \sec \theta d\theta[/tex]

    Using that substitution, I got it down to:

    [tex]\int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sqrt{\tan^2 \theta+1}} = \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sec \theta}[/tex]

    I have no clue how this is going to get to the answer. Could someone please help? Thanks.
     
  2. jcsd
  3. Jul 22, 2008 #2

    jamesrc

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    Gold Member

    What about the dx term?

    [tex]
    dx=d(\tan\theta)=??
    [/tex]
     
  4. Jul 22, 2008 #3

    Dick

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    You are forgetting the dx part again. That's why your expression differs from what you are supposed to show. When you get it right, try expressing the integrand in terms of sin and cos.
     
  5. Jul 23, 2008 #4

    LHC

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    Ah, NOW I get it. Thanks to everyone for your help!
     
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