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Integrating: What Am I Doing Wrong?

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int \frac{dx}{(1+4x^2)^{3/2}}[/itex]

    3. The attempt at a solution

    [itex]\int \frac{dx}{(1+4x^2)^{3/2}}[/itex]

    Let [itex]x = \frac{1}{4}tan(u), dx = \frac{1}{4}sec^2(u)du[/itex]

    [itex]\frac{1}{4}\int{\frac{sec^2(u)du}{(sec^2(u))^{3/2}}}[/itex]

    [itex]\frac{1}{4}\int{\frac{1}{sec(u)}}du[/itex]

    [itex]\frac{1}{4}\int{cos(u)}du[/itex]

    [itex]\frac{1}{4}sin(u) + C[/itex]


    Drawing a triangle with angle u:


    [itex]tan(u) = 4x[/itex]

    Therefore:

    Opposite = [itex]4x[/itex]

    Adjacent = [itex]1[/itex]

    Hypotenuse = [itex]\sqrt{1^2 + (4x)^2} = \sqrt{1 + 16x^2}[/itex]

    [itex]sin(u) = \frac{4x}{\sqrt{1 + 16x^2}}[/itex]


    [itex]\frac{1}{4}sin(u) + C[/itex]

    [itex]\frac{1}{4}(\frac{4x}{\sqrt{1 + 16x^2}}) + C[/itex]

    [itex]\frac{x}{\sqrt{1 + 16x^2}} + C[/itex]

    I personally can't see what I did wrong here, but this is not the correct answer :( Both my math book and Mathematica say that the correct answer should actually be

    [itex]\frac{x}{\sqrt{1 + 4x^2}} + C[/itex]

    Any help is greatly appreciated :)
     
    Last edited by a moderator: Jun 19, 2012
  2. jcsd
  3. Jun 19, 2012 #2
    Nevermind, I got it.

    The substitution I used, [itex]x = \frac{1}{4}tan(u)[/itex] does not produce the result I wanted because the [itex]\frac{1}{4}[/itex] would be squared also. Instead, I made the substitution [itex]x = \frac{1}{2}tan(u)[/itex] and that led me to the correct answer.
     
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