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Math Help - Integration

  1. #1
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    Integration

    Hi,

    \text{Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)

    \int \frac{1}{2\cos^2 \theta - 1} = \frac{1}{2} (\log(\sin(\theta)  + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K

    \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K

    F|_a^b\Rightarrow F |_0^\frac{\pi}{6}

    = \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) +  \cos\bigg(\frac{\pi}{6}\bigg)\bigg) -  \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) -  \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -<br />
   \frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) +  \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) -  \sin\bigg(0\bigg)\bigg)\bigg)

    = \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)
    Last edited by Hellbent; November 1st 2010 at 07:04 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    Your last line should actually look like:

    \dfrac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2}\bigg) - \dfrac{1}{2}\log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)

    \dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b  igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)

    now use: \log(A)-\log(B)= \log\dfrac{A}{B}.

    Then rationalize the denominator and simplify.
    Last edited by harish21; November 1st 2010 at 07:02 PM.
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  3. #3
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    Quote Originally Posted by harish21 View Post
    Your last line should actually look like:

    \dfrac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2}\bigg) - \dfrac{1}{2}\log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)

    \dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b  igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)

    now use: \log(A)-\log(B)= \log\dfrac{A}{B}.

    Then rationalize the denominator and simplify.
    \dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b  igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)

    \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)

    \log \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

    Final answer: incorrect anyway.
    Last edited by Hellbent; November 1st 2010 at 07:38 PM.
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  4. #4
    MHF Contributor harish21's Avatar
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    NO..

    \dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b  igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)

    = \dfrac{1}{2} \log \bigg(\dfrac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}}\bigg)

    =\dfrac{1}{2} \log \dfrac{\sqrt{3}+1}{\sqrt{3}-1}
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  5. #5
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    I wasn't done. Even if my answer was going to be incorrect. I'm not sure of the reason for the slow connection.
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  6. #6
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    Quote Originally Posted by Hellbent View Post
    Hi,

    \text{Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)

    \int \frac{1}{2\cos^2 \theta - 1} = \frac{1}{2} (\log(\sin(\theta)  + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K

    \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K

    F|_a^b\Rightarrow F |_0^\frac{\pi}{6}

    = \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) +  \cos\bigg(\frac{\pi}{6}\bigg)\bigg) -  \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) -  \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -<br />
   \frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) +  \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) -  \sin\bigg(0\bigg)\bigg)\bigg)

    = \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)
    This may help

    2\cos(\theta)-1=\cos(2\theta)

    This gives the integral

    \int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac  {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2}  \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Hellbent View Post
    I wasn't done. Even if my answer was going to be incorrect. I'm not sure of the reason for the slow connection.
    OKay.

    Yes, MHF is kind of slow. Most users are experiencing the same problem.

    However, I think you should be able to proceed from here:

    \dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)
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  8. #8
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    \dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)
    Proceed to where? Ha! I thought this was done, even though it doesn't resemble this \frac{1}{2}\ln(2 + \sqrt{3})
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  9. #9
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    Quote Originally Posted by TheEmptySet View Post
    This may help

    2\cos(\theta)-1=\cos(2\theta)

    This gives the integral

    \int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac  {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2}  \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|
    It doesn't help. How does the substitution help? My already poor understanding is being muddled. I don't see how it gives the integral...

    \int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac  {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2}  \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|


    \ln|2+\sqrt{3}|
    \text{Is the above the same as}\, \frac{\pi\,sec(2\theta)}{6}\text{?}
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  10. #10
    Behold, the power of SARDINES!
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    Quote Originally Posted by Hellbent View Post
    It doesn't help. How does the substitution help? My already poor understanding is being muddled. I don't see how it gives the integral...

    \int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac  {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2}  \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|


    \ln|2+\sqrt{3}|
    \text{Is the above the same as}\, \frac{\pi\,sec(2\theta)}{6}\text{?}
    From trigonomerty we have the identities that

    2\cos^2(\theta)-1=\cos(2\theta)

    We also have the identity that

    \frac{1}{\cos(2\theta)}=\sec(2\theta)

    This gives

    \frac{1}{2\cos(\theta)-1}=\frac{1}{\cos(2\theta)}=\sec(2\theta)

    Now we need to integrate

    \int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta

    FYI most people just memorize this anti-derivative but here is the substitution to work it out. We will do it in two steps. first let
    w=2\theta \implies dw=2\theta


    \frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sec(w)dw

    Now here is the part most folks memorize.

    let \displaystyle u=\sec(w)+\tan(w) \implies du=\sec(w)\tan(w)+\sec^2(w)dw=\sec(w)(\tan(w)+\sec  (w))dw=\sec(w)udw

    This gives that

    \displaystyle \frac{du}{u}=\sec(w)dw now the integral becomes

    \displaystyle \frac{1}{2}\int \frac{du}{u}=\ln|u|=\ln|\sec(w)+\tan(w)| \bigg|_{0}^{\frac{\pi}{3}}=\ln|2+\sqrt{3}|
    Last edited by TheEmptySet; November 1st 2010 at 08:31 PM. Reason: typo
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  11. #11
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    From trigonomerty we have the identities that

    2\cos(\theta)-1=\cos(2\theta)
    Correction: It might be a typo, but

    That should be 2\cos^{2}(\theta)-1=\cos(2\theta)
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  12. #12
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    What's the relationship between these two: \dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg) and \ln|2+\sqrt{3}|?
    Last edited by Hellbent; November 1st 2010 at 08:48 PM.
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  13. #13
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Hellbent View Post
    What's the relationship between these two: \dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg) and \ln|2+\sqrt{3}|?
    \dfrac{\sqrt{3}+1}{\sqrt{3}-1} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}

    = \dfrac{3+2\sqrt{3}+1}{3-1} = \dfrac{4+2\sqrt{3}}{2} = 2+\sqrt{3}

    therefore,

    \dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg) = \dfrac{1}{2} log( 2+\sqrt{3})
    Last edited by harish21; November 1st 2010 at 10:53 PM.
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  14. #14
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    Argh! I know I'm tremendously slow, but now that I see it, it's rather trvial.
    Thinking of it my rationalization was different and wrong. I used the incorrect conjugate.

    Ah, is TheEmpySet's method another approach or was he just hinting at what is directly above this post?

    Thanks for your patience.
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  15. #15
    MHF Contributor harish21's Avatar
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    The process shown by TheEmptySet is way quicker than what you had done to integrate..Its different from the one that you had followed.
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