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Trig substitution

  1. Oct 8, 2007 #1

    tony873004

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    [SOLVED] trig substitution

    This is from the class notes. Evaluate the integral:
    [tex]
    \int_{}^{} {\frac{x}{{\sqrt {3 - x^4 } }}dx}
    [/tex]

    [tex]
    \begin{array}{l}
    u = x^2 ,\,\,du = 2x\,dx\,\, \Leftrightarrow \,\,dx = \frac{{du}}{{2x}} \\
    \\
    \int_{}^{} {\frac{{x^1 }}{{\sqrt {3 - x^4 } }}dx} = \int_{}^{} {\frac{x}{{2x\sqrt {3 - u^2 } }}du} = \int_{}^{} {\frac{1}{{2\sqrt {3 - u^2 } }}du } \\
    \end{array}
    [/tex]

    The next step I would want to do using trig substitution is
    [tex]
    \begin{array}{l}
    a = \sqrt 3 ,\,x = a\sin \theta = \sqrt 3 \sin \theta \\
    \\
    \frac{1}{2}\int_{}^{} {\frac{1}{{\sqrt {\sqrt 3 ^2 - \sqrt 3 \sin \theta } }}du} = \frac{1}{2}\int_{}^{} {\frac{1}{{\sqrt {3 - \sqrt 3 \sin \theta } }}du} \\
    \end{array}
    [/tex]

    But the next step in the example is:
    [tex]
    \int_{}^{} {\frac{1}{{2\sqrt {3 - u^2 } }}du = } \frac{1}{2}\sin ^{ - 1} \frac{u}{{\sqrt 3 }} + C
    [/tex]
    How did he get this? Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2007 #2

    cristo

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    [tex]\int\frac{du}{2\sqrt{3-u^2}}=\frac{1}{2}\int\frac{du}{\sqrt{3}\sqrt{1-(u/ \sqrt{3})^2}}[/tex] then let y=u/sqrt(3) hence dy=du/sqrt(3), so we have [tex]\frac{1}{2}\int\frac{dy\sqrt{3}}{\sqrt{3}\sqrt{1-y^2}}=\frac{1}{2}\int\frac{dy}{\sqrt{1-y^2}}[/tex] which is a standard integral.
     
  4. Oct 8, 2007 #3

    tony873004

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    Thanks for the reply. But I don't see how you did this:
    [tex]\int\frac{du}{2\sqrt{3-u^2}}=\frac{1}{2}\int\frac{du}{\sqrt{3}\sqrt{1-(u/ \sqrt{3})^2}}[/tex]

    I know you're right, because I picked a random value for u and plugged both formulas into my calculator. But I just don't know how you got from one to the other.
     
  5. Oct 8, 2007 #4

    cristo

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    Ok, well let's just consider the denominator: [itex]2\sqrt{3-u^2}[/itex]. This can be rewritten as [tex]2\sqrt{(\sqrt{3})^2-u^2}=2\sqrt{(\sqrt{3})^2[1-(u/\sqrt{3})^2]}[/tex] by factoring out \sqrt{3}^2. We can then take this outside the square-root sign to give [tex]2\sqrt{3}\sqrt{1-(u/\sqrt{3})^2}[/tex] as required.
     
    Last edited: Oct 8, 2007
  6. Oct 8, 2007 #5

    tony873004

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    I was about to write back and ask how you did that factoring, but now that I've stared at it for a few minutes, I see what you did.

    They expected me to come up with that on my own??

    Thanks, Christo :)
     
  7. Oct 8, 2007 #6

    cristo

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    You're welcome!
     
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