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Homework Help: Integration of trigonometric function

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{2x}{x^{2}+6x+13}[/tex]dx

    2. Relevant equations

    [tex]\int[/tex][tex]\frac{1}{u^{2}+a^{2}}[/tex]du = [tex]\frac{1}{a}[/tex]arctan([tex]\frac{u}{a}[/tex]+c

    [tex]\int[/tex][tex]\frac{du}{u}[/tex]=ln(u) + c

    3. The attempt at a solution

    u=x+3
    2xdu=2x dx
    x=u-3

    [tex]\int[/tex][tex]\frac{2x}{u^{2}+a^{2}}[/tex]du
    =[tex]\int[/tex][tex]\frac{2u-6}{u^{2}+a^{2}}[/tex]du
    =[tex]\int[/tex][tex]\frac{2u}{u^{2}+a^{2}}[/tex]du + [tex]\int[/tex][tex]\frac{-6}{u^{2}+a^{2}}[/tex]du

    This is where I get stuck because I can only figure out the second integral which turns out to be:
    -3arctan([tex]\frac{x+3}{2}[/tex]+c

    Basically for the first integral I want to know if this is allowed?

    v=u[tex]^{2}[/tex]+a[tex]^{2}[/tex]
    dv=2u du

    [tex]\int[/tex][tex]\frac{dv}{v}[/tex]=ln(v)

    ln(v)=ln(u[tex]^{2}+a^{2}[/tex])=ln((x+3)[tex]^{2}[/tex]+4)+c

    So my answer would be the following, which agrees with the book:

    ln(x[tex]^{2}[/tex]+6x+13)-3arctan([tex]\frac{u}{a}[/tex]+c
     
  2. jcsd
  3. Sep 2, 2010 #2
    Without checking over your work (which I don't have time to do), I can say that yes, that is allowed. You can always substitute as much as you want so long as you give the answer in the desired form in the end.
     
  4. Sep 3, 2010 #3
    just remeber
    [tex]\int[/tex][tex]\frac{f'(x)}{f(x)}dx=lnf(x)[/tex]
     
  5. Sep 3, 2010 #4

    Mark44

    Staff: Mentor

    Don't forget the absolute value and the constant of integration.
    [tex]\int\frac{f'(x)}{f(x)}dx=ln|f(x)| + C[/tex]
     
  6. Sep 4, 2010 #5
    mehh ,whatever
     
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