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Stuck again still trig sub

  1. Jun 2, 2005 #1
    The question i'm working on is "use the appropriate Trigonometric substitution to evaluate the following integral: [tex]\int \frac{x^2}{(4x^2-49)^\frac{3}{2}}dx[/tex] " Now the first thing i did was make it so that it had a root in it by making it [tex]\int \frac{x^2}{(4x^2-49)(4x^2-49)^\frac{1}{2}}dx[/tex]
    this then allows me to make the substitution of [tex]x=\frac{7sec\theta}{2}[/tex] and [tex]dx=\frac{7sec\theta tan\theta}{2}d\theta[/tex].

    making the substitution i get [tex]\int\frac{(49)(sec^2\theta) * (7)(tan\theta) (sec\theta)}{(4)(2)(49)(tan^2\theta)(7)(tan\theta)}d\theta[/tex] which simplifies to [tex]\int\frac{(sec^2\theta)(tan\theta)(sec\theta)}{(8)(tan^3\theta)}d\theta = \frac{1}{8}\int\frac{sec^3\theta}{tan^2\theta}d\theta[/tex] from there i've tired all sorts of different things, messed around in all the ways i know to get it so that i can integrate, but i just dont know where to go from there.
     
  2. jcsd
  3. Jun 2, 2005 #2

    OlderDan

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    I think this might help, but I think I still see some partial fractions in your future

    [tex]\frac{sec^3\theta}{tan^2\theta} = \frac{sec \theta}{sin^2\theta} = \frac{cos \theta}{cos^2\theta sin^2\theta} = \frac{cos \theta}{(1 - sin^2\theta) sin^2\theta} [/tex]
     
    Last edited: Jun 2, 2005
  4. Jun 2, 2005 #3

    dextercioby

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    [tex] I=\int \frac{x^{2}}{\left(4x^{2}-49\right)^{3/2}} \ dx [/tex] (1)

    Write it like that

    [tex] I=\frac{1}{49^{3/2}} \int\frac{x^{2}}{\left[\left(\frac{2x}{7}\right)^{2}-1\right]^{3/2}} \ dx [/tex] (2)

    and make the sub

    [tex] \frac{2x}{7}=\cosh t [/tex] (3)

    under which the integration element behaves

    [tex] 2\frac{dx}{7}=\sinh t \ dt [/tex] (4)

    The transformed integral is

    [tex] I=\frac{1}{7^{3}}\frac{7^{3}}{2^{3}}\int \coth^{2}t \ dt=\frac{1}{8}\left(\int \frac{\sinh^{2}t +1}{\sinh^{2}t} \ dt \right)=\frac{1}{8}\left(t-\coth t\right) +\mathcal{C} [/tex] (5)

    Reverse the substitution #3.

    Daniel.
     
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