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Math Help - integrating sech^(-1) x

  1. #1
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    integrating sech^(-1) x

    the question is
    show

     <br />
\int_\frac{4}{5}^1 sech^{-1} x =2tan^{-1} 2- \frac{\pi}{2}-\frac{4ln2}{5}<br />

    i cant get the first term, i end up with sin^{-1}(\frac{4}{5})

    i proceeded by parts with

     <br />
u=arsech x =ln(\frac{1+\sqrt{1-x^2}}{x}) u'= \frac{-1}{x\sqrt{1-x^2}}<br />
    v'=1 dv=x

    so im left with integrating \frac{1}{\sqrt{1-x^2}}

    which gives me sin^{-1} x

    ive put the values 2artan(2) and arcsin(4/5) in calculator and they are not the same so not sure where im going wrong here.



    ALSO as im mentioning arsech my books gives the formula for the log form as _-^+ ln(\frac{1+\sqrt{1-x^2}}{x})

    how can this be? for the other inverse hyperbolics they pick the positive, why take both for arsech x? how do i work out arsech (3) say? how do i know whether to pick the = or the - value?
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by jiboom View Post
    the question is
    show

     <br />
\int_\frac{4}{5}^1 sech^{-1} x =2tan^{-1} 2- \frac{\pi}{2}-\frac{4ln2}{5}<br />

    i cant get the first term, i end up with sin^{-1}(\frac{4}{5})

    i proceeded by parts with

     <br />
u=arsech x =ln(\frac{1+\sqrt{1-x^2}}{x}) u'= \frac{-1}{x\sqrt{1-x^2}}<br />
    v'=1 dv=x

    so im left with integrating \frac{1}{\sqrt{1-x^2}}

    which gives me sin^{-1} x

    ive put the values 2artan(2) and arcsin(4/5) in calculator and they are not the same so not sure where im going wrong here.



    ALSO as im mentioning arsech my books gives the formula for the log form as _-^+ ln(\frac{1+\sqrt{1-x^2}}{x})

    how can this be? for the other inverse hyperbolics they pick the positive, why take both for arsech x? how do i work out arsech (3) say? how do i know whether to pick the = or the - value?

    \int_{\frac{4}{5}}^{1} sech^{-1}x\; dx

    dv = dx \Rightarrow v=x

    u=sech^{-1} \Rightarrow du = \frac{-1}{x\sqrt{1-x^2}}

    \int_{\frac{4}{5}}^{1} sech^{-1}x\; dx = x \; sech^{-1} x + \int \frac{1}{\sqrt{1-x^2}} \;dx

    \int_{\frac{4}{5}}^{1} sech^{-1}x\; dx = x \; sech^{-1} x + \sin ^{-1} x

    \int_{\frac{4}{5}}^{1} sech^{-1}x\; dx = x\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right) + \sin ^{-1} x

    look to the picture
    integrating sech^(-1) x-nice.jpg

    \sin^{-1} x = \tan ^{-1} \frac{x}{\sqrt{1-x^2}}=u
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  3. #3
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    hi, thanks for the reply.

    i still dont evaluate it to the solution.

     <br />
  ^1 _ {\frac{4}{5}}  [\tan ^{-1} \frac{x}{\sqrt {1-x^2}}]<br />

    = -\tan^{-1} \frac{4}{3}+\frac{\pi}{2}
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  4. #4
    MHF Contributor Amer's Avatar
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    I will show you that
    - 2\tan ^{-1} 2 = \tan ^{-1} \frac{4}{3}

    if I show that

    \tan (-2 \tan ^{-1} 2) = \frac{4}{3}

    I finished

    \tan 2x = \frac{2\tan x }{1- \tan^2 x}

    -\tan 2 \tan ^{-1} (2) = -\frac{2 \tan ( \tan ^{-1} 2)}{1- ( \tan ( \tan ^{-1} 2))^2} = -\frac{4}{1-4} = \frac{4}{3}

    end
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  5. #5
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    nice,thank-you. Im not sure why my calc gave me different answers All that is left now is to explain the differen signs. Any ideas ? As i put in my OP my book uses + and - ln{...} for arsech, and + and - for its derivative so it looks like they have chosen the other sign. Can you see why ?
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  6. #6
    MHF Contributor Amer's Avatar
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    first let us find
    arcsech \; x equal what

    arcsech \; x = u \Rightarrow x = sech u

    sech u = \frac{2}{e^{u}-e^{-u}}

    xe^{u} + xe^{-u} - 2 = 0

    xe^{2u} - 2e^{u} + x = 0

    e^u = \frac{-(-2) \mp \sqrt{4 - 4x^2}}{2x}

     u = \ln \left( \frac{1 \mp \sqrt{1- x^2}}{x} \right) = arcsech\; x

    we have two values for arcsech\; x

    if we choose arcsech\; x =\ln \left( \frac{1- \sqrt{1-x^2}}{x} \right)

    what we will get

    \int_{\frac{4}{5}}^{1} arcsech\; x \;dx

    u = \ln \left( \frac{1- \sqrt{1-x^2}}{x} \right)

    u' = \frac{1}{x\sqrt{1-x^2}}

    dx =dv \Rightarrow v=x

    \int_{\frac{4}{5}}^{1} arcsech \;x\; dx = x \ln \left( \frac{1- \sqrt{1-x^2}}{x} \right) - \int \frac{1}{\sqrt{1-x^2}} dx

    \int_{\frac{4}{5}}^{1} arcsech\; x\; dx =x \ln \left( \frac{1- \sqrt{1-x^2}}{x} \right) - \sin ^{-1} x

    x \ln \left( \frac{1- \sqrt{1-x^2}}{x} \right) \mid_{\frac{4}{5}}^{1} = 0 - (4/5 \ln (2/4)) = 4/5 ln(2) positive sign

    \sin ^{-1} x \mid_{\frac{4}{5}}^{1} = \frac{\pi}{2} - \sin ^{-1} 4/5

    \int_{\frac{4}{5}}^{1} arcsech\; x\; dx = \frac{4\ln(2)}{5} - \left(\frac{\pi}{2} - \sin ^{-1} \frac{4}{5} \right)

    if we want to see what is the realtion between \sin ^{-1} 4/5 and 2\tan ^{-1} 2

    \tan ^{-1} 2 = \sin ^{-1} \frac{2}{\sqrt{5}}

    2 \sin^{-1} \frac{2}{\sqrt{5}}  this is an angle and it is equal to 2\cos ^{-1} \frac{1}{\sqrt{5}}
    if it is not clear to you draw a triangle

    \sin 2 \left(\sin^{-1} \frac{2}{\sqrt{5}} \right)= 2 \sin \left(\sin^{-1} \frac{2}{\sqrt{5}}  \right) \cos \left(\sin^{-1} \frac{2}{\sqrt{5}}  \right)

    \sin 2 \left(\sin^{-1} \frac{2}{\sqrt{5}} \right)= 2 \sin \left(\sin^{-1} \frac{2}{\sqrt{5}}  \right) \cos \left(\cos^{-1} \frac{1}{\sqrt{5}}  \right)

    \sin 2 \left(\sin^{-1} \frac{2}{\sqrt{5}} \right)= 2 \frac{2}{\sqrt{5}} \frac{1}{\sqrt{5}} = \frac{4}{5}

    take sin ^{-1} inverse sin function for the both sides

    2 \left(\sin^{-1} \frac{2}{\sqrt{5}} \right) = \sin ^{-1}\frac{4}{5}

    \int_{\frac{4}{5}}^{1} arcsech\; x\; dx = \frac{4\ln(2)}{5} - \frac{\pi}{2} +2 \tan ^{-1} (2)

    different answer why, I think the problem is in the angles since \sin a = \sin (a+ 2\pi)

    I said I think
    I wish you understand why and see how it comes
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  7. #7
    MHF Contributor Amer's Avatar
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    see these two links

    link 1
    link 2

    I think we should use \ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)

    since it is same as arcsech x graph exactly but the other form is not like arcsech x
    Last edited by Amer; May 1st 2010 at 06:01 AM.
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  8. #8
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    thanks for taking the time to look at this. As everything points to the signs being wrong in the question, wolfram does the integral of arsech to the positive value, ill amend the question.
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  9. #9
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    Quote Originally Posted by Amer View Post
    I will show you that
    - 2\tan ^{-1} 2 = \tan ^{-1} \frac{4}{3}

    if I show that

    \tan (-2 \tan ^{-1} 2) = \frac{4}{3}

    I finished

    \tan 2x = \frac{2\tan x }{1- \tan^2 x}

    -\tan 2 \tan ^{-1} (2) = -\frac{2 \tan ( \tan ^{-1} 2)}{1- ( \tan ( \tan ^{-1} 2))^2} = -\frac{4}{1-4} = \frac{4}{3}

    end
    having trouble with this. wofram is saying this is not true.

    It gives tan^(-1)[4/3]=0.92729521 = sin^(-1) [4/5]
    arctan&#x5b;4&#x2f;3&#x5d; - Wolfram|Alpha

    and

    tan^(-1) [2]= 1.107148715 = sin^(-1)[2/rt(5)]

    arctan&#x5b;2&#x5d; - Wolfram|Alpha

    something is going awry here. Either im entering the expressions wrongly or they are not equal??!
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  10. #10
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    the problem is not to show artan(4/3)=2tan(2) as it is not true but to show

    pi/2-artan(4/3)=2artan(2)-pi/2 which is true

    any clues how to do thisV NICELY?

    I can get values for the artans and get 2artan(2)+artan(4/3)=126.86+53.13=180 [degrees]
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