1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrals (complex) please help

  1. Feb 1, 2006 #1
    im stuck at this integral where u=t

    please help

    [​IMG]

    i know the integral of (du)/(u^2+1) equals arctan(u), i think that might have something to do with it



    thanks
     
  2. jcsd
  3. Feb 1, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    But you don't HAVE "u2+ 1", you have "u2+ 3". So try [itex]u^2+ 3= 3(\frac{1}{3}u^2+ 1)[/itex] and let [itex]v= \frac{1}{\sqrt{3}}u[/itex]. Then [itex]dv= \frac{1}{\sqrt{3}}du[/itex] so that [itex]du= \sqrt{3}dv[/itex]. Of course, [itex]u^2+ 3= 3(\frac{1}{3}u^2+1)= 3(v^2+ 1). When u= 1, [itex]v= \frac{1}{\sqrt{3}}[/itex] and when [itex]u= e^t[/itex], [itex]v= \frac{1}{3}e^t[/itex]. Your integral becomes:
    [tex]\frac{\sqrt{3}}{3}\int_\frac{1}{\sqrt{3}}^{\frac{1}{\sqrt{3}}e^t}\frac{dv}{v^2+1}[/tex]

    Yes, the integral of that is [itex]\frac{\sqrt{3}}{3}arctan(v)[/itex] evaluated between the two limits.
     
    Last edited: Feb 1, 2006
  4. Feb 1, 2006 #3
    whoaa i dont think im that advanced yet, but thanks man. ill try to use it

    :cool:
     
  5. Feb 1, 2006 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "Advanced"?? The most advanced calculus used was the arctan integral. The rest is all algebra.
     
  6. Feb 1, 2006 #5
    would this be equal to ((3^.5)/3) x ((2Theta)/6)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integrals (complex) please help
Loading...