Arctan integral

  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?


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

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

    [tex]6* \frac{1}{3(4/3u^2+1)}[/tex]
  2. jcsd
  3. George Jones

    George Jones 6,474
    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. something was wrong in step 4.
    [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. VietDao29

    VietDao29 1,422
    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. 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]
    [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. VietDao29

    VietDao29 1,422
    Homework Helper

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

Have something to add?