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Math Help - Integrating using Trig Substitution

  1. #1
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    Integrating using Trig Substitution

    The problem is

    (x^4) / (sqrt x^10 -2)


    I thought I could use U- Substitution first and work it out like that...but that doesn't pan out. Help please this one has me stumped.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    The problem is

    (x^4) / (sqrt x^10 -2)


    I thought I could use U- Substitution first and work it out like that...but that doesn't pan out. Help please this one has me stumped.
    Let x^5=\varphi

    so d\varphi=5x^4dx

    Giving us

    \frac{-1}{5}\int\frac{d\varphi}{2-\varphi^2}

    Now let \varphi=\sqrt{2}\tanh(\theta)

    and go from there

    EDIT:

    I am sorry I misread it, it should be

    \frac{1}{5}\int\frac{d\varphi}{\sqrt{\varphi^2-2}}

    Now let \varphi=\sqrt{2}\sec(\theta)
    Last edited by Mathstud28; June 14th 2008 at 03:19 PM.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Let x^5=\varphi

    so d\varphi=5x^4dx

    Giving us

    \frac{-1}{5}\int\frac{d\varphi}{2-\varphi^2}

    Now let \varphi=\sqrt{2}\tanh(\theta)

    and go from there

    EDIT:

    I am sorry I misread it, it should be

    \frac{1}{5}\int\frac{d\varphi}{\sqrt{\varphi^2-2}}

    Now let \varphi=\sqrt{2}\sec(\theta)


    OK I worked it out from where you left off and eventually I get to a point where it looks like this

    (1/5) integrate (SecTheta)

    When i integrate that, I end up with

    (1/5) ln(abs(secTheta + tan Theta))

    What am i supposed to do from here?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    OK I worked it out from where you left off and eventually I get to a point where it looks like this

    (1/5) integrate (SecTheta)

    When i integrate that, I end up with

    (1/5) ln(abs(secTheta + tan Theta))

    What am i supposed to do from here?
    \sqrt{2\sec(\theta)^2-2}=\sqrt{2}\color{red}{\tan(\theta)}
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  5. #5
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    Ok made the correction for tanTheta and ended up with having to integrate secTheta which my final line is


    (1/5) ln l sectheta + tantheta l + C

    Now looking at my teachers answer, this is far from correct but i'm not sure as to how to proceed from here.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Ok made the correction for tanTheta and ended up with having to integrate secTheta which my final line is


    (1/5) ln l sectheta + tantheta l + C

    Now looking at my teachers answer, this is far from correct but i'm not sure as to how to proceed from here.
    \frac{1}{5\sqrt{2}}\int\sec(\theta)d\theta=\frac{1  }{5\sqrt{2}}\ln|\sec(\theta)+\tan(\theta)|

    x^5=\varphi=\sqrt{2}\sec(\theta)\Rightarrow{arcsec  \bigg(\frac{x^5}{\sqrt{2}}<br />
\bigg)=\theta}

    So now back sub
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  7. #7
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    MmMm.... how are you getting those values? I remember the teacher saying something about drawing a triangle to figure out how to get everything back in terms of x. I think this is the part but I didn't understand it too well. Could you show me?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    MmMm.... how are you getting those values? I remember the teacher saying something about drawing a triangle to figure out how to get everything back in terms of x. I think this is the part but I didn't understand it too well. Could you show me?
    Obviously

    \sec\bigg(arcsec\bigg(\frac{x^5}{\sqrt{2}}\bigg)\b  igg)=\frac{x^5}{\sqrt{2}}

    and

    \tan\bigg(arcsec\bigg(\frac{x^5}{\sqrt{2}}\bigg)\b  igg)=\frac{\sqrt{x^{10}-2}}{\sqrt{2}}
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