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Is this a good substitution that will work

  1. Sep 13, 2007 #1

    rock.freak667

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    Homework Helper

    1. The problem statement, all variables and given/known data
    Prove [tex]\int_0^{1} \frac{1}{\sqrt{x^2+6x+25}} = ln(\frac{1+\sqrt{2}}{2})[/tex]


    2. Relevant equations



    3. The attempt at a solution

    [tex]\int_0^{1} \frac{1}{\sqrt{x^2+6x+25}}

    = \int_0^{1} \frac{1}{\sqrt{(x+3)^2+16}}[/tex]

    Let [tex]x+3=4tan\theta[/tex] so that [tex]dx=4sec^2\theta d\theta[/tex]

    and so the problem becomes

    [tex]\int \frac{4sec^2\theta}{\sqrt{16tan^2\theta+16}} d\theta[/tex]

    giving [tex]\int sec\theta d\theta = ln|sec\theta + tan\theta|+ K[/tex]
     
    Last edited: Sep 13, 2007
  2. jcsd
  3. Sep 14, 2007 #2

    danago

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    Gold Member

    Isolate theta in the substitution you made i.e. [tex]\theta=arctan\frac{x+3}{4}[/tex]. From there, you should be able to evaluate the definite integral, and come to the required solution.
     
  4. Sep 14, 2007 #3

    Gib Z

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    You are correct so far, now just change the bounds accordingly to your substitutions. The first bound, x=1, so put that into your subsitution, tan theta = 1, ie theta = pi/4. Do the same for the other bound, and evaluate from your last line.
     
  5. Sep 14, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It is not absolutely necessary to let
    [tex]\theta=arctan\frac{x+3}{4}[/tex]
    (and then use trig identities). Imagine a right triangle with one angle [itex]\theta[/itex] since you know
    [tex]tan(\theta)= \frac{x+3}{4}[/tex],
    the triangle has "opposite side" of length x+3 and "near side" of 4.

    By the Pythagorean theorem, the square of the hypotenuse is [itex](x+3)^2+ 16= x^2+ 6x+ 25[/itex].

    Then [itex]sec(\theta)[/itex], hypotenuse over near side is [tex]\frac{x^2+ 6x+ 25}{4}[/tex]
    and
    [tex]tan(\theta)[/tex]
    is, of course,
    [tex]\frac{x+3}{4}[/itex].
     
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