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Integral help

  1. Dec 10, 2003 #1
    heres a funky one I can't do
    [tex]
    \int \frac{6t \,dt}{\sqrt{1-2t^2}}
    [/tex]

    here's what I've done
    I changed it so it looks like this:
    [tex]
    \int 6t(1-2t^2)^{-1/2} dt
    [/tex]
    Then I substituted u for 1-2t^2 and got du= -4t*dt
    Where do I go from here?
     
  2. jcsd
  3. Dec 10, 2003 #2
    ...

    Take 6 as 4 * 1.5 and put [tex](1-2t^2) = u [/tex] so that [tex]4tdt[/tex] becomes [tex]du[/tex]. Then you can proceed normally. The integral now contains [tex]1.5*u^{-1/2}du[/tex].

    Sridhar
     
    Last edited: Dec 11, 2003
  4. Dec 11, 2003 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Rather than "seeing" that 6= 4*1.5,

    Another way to look at this is: taking u to be 1-2t2 so that du= -4tdt, then tdt= -(1/4)du and 6tdt/√(1-2t2) becomes
    6(-1/4)du/u1/2= (-3/2)u-1/2du
     
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