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!

Arctan integral

  1. Sep 18, 2006 #1
    Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4)


    according to the book the answer is sqrt(3) * arctan(2u/sqrt(3))
    but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function.

    The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra?

    1
    [tex]
    \frac{3/2}{u^2+3/4}
    [/tex]

    2
    [tex]3/2 * \frac{1}{(4u^2+3)/4}[/tex]

    3
    [tex]3/2*4 * \frac{1}{4u^2+3}[/tex]

    4
    [tex]6* \frac{1}{3(4/3u^2+1)}[/tex]
     
  2. jcsd
  3. Sep 18, 2006 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    Careful. You know that

    [tex]\int{ \frac{1}{v^2 + 1} dv = arctan(v) + c.[/tex]

    So, for your question, what is [itex]v[/itex]?
     
  4. Sep 20, 2006 #3
    something was wrong in step 4.
    becuse
    [tex]\int { \frac{1} {px^2+1} dx = \frac{1} {\sqrt{p}} arctan(\sqrt{p} x) +c [/tex]
    here p > 0.

    thus,the correct answer is
    [tex]\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c[/tex]
     
  5. Sep 21, 2006 #4

    VietDao29

    User Avatar
    Homework Helper

    Fine still step 4. :)
    You know that:
    [tex]\int \frac{du}{u ^ 2 + 1} = \arctan (u) + C[/tex], right?
    And now, you've gotten to:
    [tex]\int \frac{6 dx}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 dx}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}[/tex]
    Now, to find the anti-derivative of this expression, you can use the substitution: [tex]v = \frac{2}{\sqrt{3}} u[/tex].
    Can you go from here? :)
     
  6. Sep 21, 2006 #5
    OK,let`s see it

    [tex]\int \frac{6 du}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}[/tex]
    and now we use the substitution: [tex]p = \frac{2}{\sqrt{3}} u[/tex]
    then we get this equation: [tex]u =\frac{\sqrt{3}}{2} p[/tex]
    so:
    [tex]\int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}=\int \frac{2}{p^2+1}d(\frac{\sqrt{3}}{2} p) = \sqrt{3}actan(p) + c[/tex]
    finally we get the answer:
    [tex]\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c[/tex]
     
  7. Sep 21, 2006 #6

    VietDao29

    User Avatar
    Homework Helper

    Yup, it's correct, congratulations. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Arctan integral
Loading...