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More Trig Subsitution

  1. Mar 14, 2007 #1
    Aye gents, I've been working on a problem that doesn't have a solution at the back of the book and would be grateful if you'd compare my answer with yours.

    1. The problem statement, all variables and given/known data

    [tex]\int\frac{t^5}{\sqrt{t^2 + 2}}\ dt[/tex]


    2. Relevant equations

    [tex]t = \sqrt{2}\tan\theta[/tex]

    [tex]dt = \sqrt{2}\sec^2\theta\ d\theta[/tex]


    3. The attempt at a solution

    [tex]\int\frac{(\sqrt{2}\tan\theta)^5}{\sqrt{(\sqrt{2}\tan\theta)^2 + 2}}\ \sqrt{2}\sec^2\theta\ d\theta[/tex]

    [tex]\int\frac{8\tan^5\theta\sec^2\theta}{\sqrt{2\tan^2\theta + 2}\ d\theta}[/tex]

    [tex]8\int\frac{\tan^5\theta\sec^2\theta}{\sqrt{2}\sec\theta}\ d\theta[/tex]

    [tex]\frac{8}{\sqrt{2}}\int\tan^5\theta\sec\theta\ d\theta[/tex]

    [tex]\frac{8}{\sqrt{2}}\int\frac{\sin^5\theta}{\cos^6\theta}\ d\theta[/tex]

    [tex]\frac{8}{\sqrt{2}}\int\frac{[(1 - \cos^2\theta)(1 - \cos^2\theta)]}{\cos^6\theta}\ \sin\theta\ d\theta[/tex]

    [tex]\frac{8}{\sqrt{2}}\int[\frac{1}{\cos^6\theta} - \frac{2}{\cos^4\theta} + \frac{1}{\cos^2\theta}]\ \sin\theta\ d\theta[/tex]

    [tex]\frac{8}{\sqrt{2}}\ [\frac{-1}{7}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^7) + \frac{2}{5}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^5) - \frac{1}{3}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^3)]\ +\ C[/tex]


    Thanks.
     
    Last edited: Mar 14, 2007
  2. jcsd
  3. Mar 14, 2007 #2

    Dick

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    Where the heck are all of the logs coming from??? Besides, you don't even need to do a trig substitution for this one. Just put u=t^2+2.
     
  4. Mar 14, 2007 #3

    VietDao29

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    As Dick has pointed out, you are all correct from the start, except the last part, why should ln appear in that expression?
    It should be continued by letting u = cos x like this:

    [tex]...= - \frac{8}{\sqrt{2}} \int \left( \frac{1}{\cos ^ 6 \theta} - \frac{2}{\cos ^ 4 \theta} + \frac{1}{\cos ^ 2 \theta} \right) d(\cos (\theta)) = \frac{8}{\sqrt{2}} \left( \frac{1}{5 \cos ^ 5 \theta} - \frac{2}{3 \cos ^ 3 \theta} + \frac{1}{\cos \theta} \right) + C[/tex]

    Now, change everything back to t.

    [tex]t = \sqrt{2} \tan \theta , \quad \theta \in \left[ -\frac{\pi}{2} , \ \frac{\pi}{2} \right][/tex]

    [tex]\Rightarrow \tan \theta = \frac{t}{\sqrt{2}}[/tex]

    Note that [tex]\theta \in \left[ -\frac{\pi}{2} , \ \frac{\pi}{2} \right] \Rightarrow \cos \theta \geq 0[/tex], so:

    [tex]\cos \theta = \sqrt{\frac{1}{1 + \tan ^ 2 \theta}} = \sqrt{\frac{2}{2 + t ^ 2}}[/tex]

    Just plug all that in and see what you get. :)

    -----------------

    Or, even faster, you can do it like this:
    [tex]\int \frac{t ^ 5}{\sqrt{t ^ 2 + 2}} dt = \frac{1}{2} \int \frac{t ^ 4}{\sqrt{t ^ 2 + 2}} d(t ^ 2 + 2)[/tex]

    [tex]= \int t ^ 4 d(\sqrt{t ^ 2 + 2}) = \int \left( (\sqrt{t ^ 2 + 2}) ^ 2 - 2 \right) ^ 2 d(\sqrt{t ^ 2 + 2}) = ...[/tex]
    Can you go from here? :)
     
    Last edited: Mar 14, 2007
  5. Mar 14, 2007 #4

    HallsofIvy

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    That looks like the hard way to me. Rewrite the integral as
    [itex]\int \frac{t^5}{\sqrt{t^2+ 1}dt= n\int \frac{(t^2)^2}{\sqrt{t^2+1} (dt)[/itex]so, with u= t2+1[/itex]] we have
    [tex]\int \frac{(u-1)^2}{u^{a+1)}du[/tex]
     
  6. Mar 15, 2007 #5
    What's wrong with having natural logs in the solution? How does one know to avoid ln when approaching this problem?

    I tried Dick's substitution of [tex]u = t^2 + 2[/tex] and didn't get very far.

    [tex]u = t^2 + 2[/tex]

    [tex]du = 2t\ dt[/tex]

    [tex]\frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t[/tex]

    Integration by parts after this step looks like something I want to avoid. Maybe I'm a bit unclear of how to use Dick's substitution?


    Thank you for the ideas, VietDao, but I'm thrown off by the suggestion...

    [tex]\frac{1}{2} \int \frac{t ^ 4}{\sqrt{t ^ 2 + 2}} d(t ^ 2 + 2)[/tex]

    Shouldn't the [tex]d(t^2 + 2)[/tex] read [tex]d(2t)[/tex] ?


    HallsofIvy, I'm having trouble understanding how the following is legal using a substitution of [tex]u = t^2 + 1[/tex]:

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


    Thanks.
     
    Last edited: Mar 15, 2007
  7. Mar 15, 2007 #6

    Gib Z

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    He never said anything was wrong with Natural logs in a solution, maybe he was pointing out that they just werent correct at all.

    Dick's suggestion should go more like this.
    You have the right set up, but forgot the dt: [tex]\frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t dt[/tex]

    Now We make the substitution, u=t^2 + 2, so the denominator is sqrt of u!,

    the 2t dt, as you correctly wrote, is du.

    t^4 is (u-2)^2.

    So you get the pleasure of doing
    [tex]\frac{1}{2}\int \frac{(u-2)^2}{\sqrt{u}} du[/tex] instead :D

    VietDao's suggestion is the same as dicks, but we just wrote the t^2+2 as u, didnt we :D

    I don't understand halls method either..assuming he left the 1/2 outside the integral out, I understand his numerator and differential but not the denominator..I thought that should be [tex]\sqrt{u+1}[/tex]..
     
  8. Mar 15, 2007 #7
    Why is ln wrong? If I have the following problem...

    [tex]\frac{1}{2}\int\frac{\sin\theta}{cos^2\theta}\ d\theta[/tex]

    my first inclination is to solve it as follows:

    [tex]\frac{1}{2}\ln(cos^2\theta) + C[/tex]


    Fill me in.
     
  9. Mar 15, 2007 #8

    Gib Z

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    [tex]\int \frac{1}{x} dx = \ln x[/tex]

    But [tex]\int \frac{1}{x^2} =-\frac{1}{x}[/tex]

    Have you seen the power rule for differentiation? Use the reverse power rule for integration, except when the power is -1, that is the only case when the reverse power rule does not work and you have to use the Natural Log function.
     
  10. Mar 15, 2007 #9

    Gib Z

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    In fact, to see why you are so obviously wrong, try differentiating your result, you won't get your integrand, thats for sure.
     
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