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Integration techniques

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data
    In each of the following cases use the given substitution in order to evaluate the given integral:

    [tex]\int\frac{2+\sqrt x}{1-\sqrt x}dx=2(4(1+\sqrt x)^\frac{3}{2}-\frac{(1-\sqrt x)^2}{2}-3In(1-\sqrt x)[/tex]




    2. Relevant equations

    For the substitution
    [tex]u=1-\sqrt x[/tex]

    [tex]x=(1-u)^2[/tex]

    3. The attempt at a solution

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

    [tex]-2\sqrt x du=dx[/tex]

    [tex]-2\sqrt(1-u)^2du=dx[/tex]


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

    [tex]-2\int 2\sqrt(1-u^2)(u^-1)du+ \sqrt(1-u^2)(u^-1)[/tex]

    Guys how come when l simplify this integration l do not get the answer correct answer which is shown in the problem statement ? Where have l done my mistake ?
     
  2. jcsd
  3. Mar 14, 2009 #2

    gabbagabbahey

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    Surely you mean [tex]\frac{du}{dx}=\frac{-1}{2\sqrt x}[/tex]...right?

    Technically, this is incorrect; since sqrt() always reteurns the positive root,

    [tex]\sqrt{(1-u)^2}=\left\{\begin{array}{lr}1-u, & 1-u \geq 0 \\ u-1, & 1-u\leq 0\end{array}[/tex]

    But it should be clear from your definition of u that [itex]u=1-\sqrt{x}\implies \sqrt{x}=1-u[/itex] always. So [itex]\sqrt{x}\neq\sqrt{(1-u)^2}[/itex] in general.


    Now you seem to be claiming [itex]\sqrt{(1-u)^2}=\sqrt{1-u^2}[/itex]....surely you didn't mean to do that!:wink:

    Use [itex]\sqrt{x}=1-u[/itex] instead.
     
  4. Mar 14, 2009 #3

    Maybe l should take a new approach since this one was a complete disaster.

    [tex]x=(1-u)^2 [/tex]

    [tex] dx= -2(1-u)du[/tex]

    [tex] dx=2u-2du[/tex]

    [tex]\int\frac{2+\sqrt (1-u)^2}{u}.(2u-2)du[/tex]

    I would like to know if l am on the right path before l take another wild goose chase:smile:
     
  5. Mar 14, 2009 #4

    gabbagabbahey

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    Again, [itex]\sqrt{(1-u)^2}[/itex] depends on whether [itex]1-u[/itex] is positive or negative.

    To avoid having to analyze the two different cases, stick to your original substitution; [itex]u=1-\sqrt{x}\implies \sqrt{x}=1-u[/itex]....
     
  6. Mar 14, 2009 #5

    Ohhh, l now get it. This means my integrand will become:


    [tex]
    \int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-2}{u}du + \int\frac{(2u-2)(1-u)}{u}du
    [/tex]
     
  7. Mar 14, 2009 #6

    gabbagabbahey

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    Actually, [tex]\int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-4}{u}du + \int\frac{(2u-2)(1-u)}{u}du[/tex]

    But there is a much easier way to simplify this:

    [tex]\int\frac{2+ (1-u)}{u}(2u-2)du=2\int\frac{(3-u)(u-1)}{u}du[/tex]

    And [itex](3-u)(u-1)=[/itex]___?
     
  8. Mar 14, 2009 #7


    Thanks very much for the help:smile:
     
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