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Homework Help: Integral is typical arctan type

  1. Apr 6, 2005 #1
    [tex]\int_{0}^{1}\frac{4}{1+x^2}[/tex]

    [tex]\int 4(1+x^2)^{-1}[/tex]

    [tex]u=1+x^2[/tex]

    [tex]du=2x[/tex]

    [tex]\frac{2du}{x}=4dx[/tex]

    [tex]2\int \frac{u^{-1}}{x} dx[/tex]

    [tex]2ln(ux)[/tex]

    [tex]=2ln(x+x^3)|_{0}^{1}[/tex]

    [tex]=2ln(1+1^3)-2ln(0+0^3)[/tex]

    I know I did something wrong, any ideas?
     
  2. jcsd
  3. Apr 6, 2005 #2

    dextercioby

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    Yes,your integral is typical [itex] \arctan [/itex] type...You should know this

    [tex] \int \frac{dx}{1+x^{2}} [/tex]

    by heart...

    Daniel.
     
  4. Apr 6, 2005 #3
    If you make a substitution, you need to arrange it such that the resulting expression does not contain the variable you substituted for.

    Here, you had

    [tex]\int 4(1+x^2)^{-1} \ dx,[/tex]

    and made the substitution

    [tex]u = 1 + x^2 \Longrightarrow du = 2x \ dx \Longrightarrow \frac{du}{2x} = dx[/tex]

    you now need to express [itex]dx[/itex] in terms of only [itex]u[/itex], ie. with no [itex]x[/itex]'s on the left side. You can do this by noting

    [tex]2x = 2\sqrt{(x^2 + 1) - 1} = 2\sqrt{u - 1}[/tex]

    so you get

    [tex] dx = \frac{du}{2\sqrt{u - 1}}[/tex]

    so your integral is

    [tex]\int \frac{4}{2u\sqrt{u-1}} \ du = \int \frac{2}{u\sqrt{u-1}} \ du[/tex]

    which is not very nice. I would advise seeing if you can find a better strategy to start with.
     
  5. Apr 6, 2005 #4

    SpaceTiger

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    When you make a substitution, you want to express everything in terms of the new variable before integrating. Try instead the substitution:

    [tex]x=tan(u)[/tex]
     
  6. Apr 6, 2005 #5
    I've never really learned about acrtan... how does tan(u) fit into this equation?
     
  7. Apr 6, 2005 #6

    dextercioby

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    If u know the substitution metod and apply it for the given function (tangent) u'll learn.

    Daniel.
     
  8. Apr 6, 2005 #7
    Subsituting [itex] x = \tan{u} \Longrightarrow dx = \sec^2 u \ du[/itex] simplifies the integral completely. Substituting these in gives

    [tex] \int \frac{4}{1+x^2} \ dx = \int \frac{4}{1 + \tan^2 u} \sec^2 u \ du[/tex]

    Then all you need to know is

    [tex]\sec^2 u - \tan^2 u = 1[/tex]

    which you can prove on your own, and I'm sure you can finish from there :wink:
     
  9. Apr 6, 2005 #8

    dextercioby

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    Whozum,you're wrong.

    Daniel.

    EDIT:Yoiu saw it & deleted the post... :rolleyes:
     
  10. Apr 6, 2005 #9
    This is a fairly straightfoward arctangent rule as others have said.

    [tex]\int \frac{4}{1 + x^2}dx = 4 \int\frac{1}{1 + x^2}dx = 4\arctan{x} + C[/tex]

    The rule for tangent inverses is [tex]\int\frac{du}{a^2+u^2} = \frac{1}{a}\arctan\frac{u}{a} + C[/tex]
     
    Last edited: Apr 6, 2005
  11. Apr 6, 2005 #10
    don't forget your constants~
     
  12. Apr 6, 2005 #11
    Well, since this was a definite integral I thought I'd let the poster apply the bounds, but yes, you are correct. I'll change them...
     
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