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Another trig. sub

  1. Feb 10, 2005 #1
    i dont know why i'm having so much problem finding a appropriate trigonometric sub for x. I can do the rest fine as long as i can find the right sub. please help me with the right trig. sub for this problem:


    [tex]\int x*sqrt(6x^2-36x+49)dx[/tex]

    here's what i done:

    [tex]\int 6x*sqrt(x^2-6x+\frac{49}{6})[/tex]

    then i did complete the square...

    [tex]\int 6x* sqrt((x-3)^2 -5/6)[/tex]

    so i came up with x = sqrt(5/6)*sec(t) as my trig. sub, but it's incorrect. can someone lend me a hand?
     
  2. jcsd
  3. Feb 10, 2005 #2
    [tex]
    \[
    \begin{gathered}
    \int_{}^{} {x\sqrt {6x^2 - 36x + 49} dx} \hfill \\
    = \int_{}^{} {x\sqrt 6 \sqrt {x^2 - 6x + \frac{{49}}
    {6}} } dx \hfill \\
    = \sqrt 6 \int_{}^{} {x\sqrt {\left( {x - 3} \right)^2 - \left( {\sqrt {\frac{5}
    {6}} } \right)^2 } } dx \hfill \\

    \hbox{Put }
    x - 3 = \sqrt {\frac{5}
    {6}} \sec \theta \hfill \\
    dx = \sqrt {\frac{5}
    {6}} \sec \theta \tan \theta d\theta \hfill \\
    \sqrt 6 \int_{}^{} {\left( {\sqrt {\frac{5}
    {6}} \sec \theta + 3} \right)} \sqrt {\frac{5}
    {6}} \tan \theta \sqrt {\frac{5}
    {6}} \sec \theta \tan \theta d\theta \hfill \\
    \hbox{I think you can do the rest (though this is not the fastest method)}
    \end{gathered}
    \]

    [/tex]
     
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