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U substitution

  1. Sep 10, 2007 #1
    First day of u subs...

    [tex]\int t \sqrt{7t^2+12}dt[/tex]

    I am assuming that u=t, but It is maiking a mess when I do that.

    Just a hint please,
    Casey
     
  2. jcsd
  3. Sep 10, 2007 #2
    your first choice on a u-substitution with a rational number should always be the entire thing under the square root.

    try u=7*t^2+12 instead
     
  4. Sep 10, 2007 #3
    let u be the radican (is that the proper term? i forget) :D
     
  5. Sep 10, 2007 #4
    do you mean like this?

    If I do this, I get [tex]\int tudt[/tex] and [tex]du=\frac{dt}{2\sqrt{7t^2+12}}[/tex] ...right?

    I think I am confused...
     
  6. Sep 10, 2007 #5
    no just the 7t^2+12

    if u=7t^2+12
    what is dt=??
     
  7. Sep 10, 2007 #6
    Oh..one sec...
     
  8. Sep 10, 2007 #7
    Brain Cramp![tex]u=7t^2+12[/tex]
    so
    [tex]du=14tdt[/tex]
    [tex]\int t u^{1/2} dt *14*\frac{1}{14}[/tex]
    [tex]=\frac{1}{14}\int \sqrt{u}* du[/tex]
    and I got it from here..
    Thanks guys,
    Casey
     
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