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Integrating arcsech x

  1. May 17, 2014 #1
    1. The problem statement, all variables and given/known data
    I was asked to prove the integral

    ##\int_{\frac{4}{5}}^{1} \textrm{arcsech}(x) =2\arctan 2-\frac{\pi}{2}-\frac{4}{5} \ln 2##


    2. Relevant equations
    Integration by parts


    3. The attempt at a solution
    Let ##u=\textrm{arcsech} (x)##
    ##\textrm{sech u}=x##
    ##\textrm{cosh u}=\frac{1}{x}##
    Differentiating implicitly,

    ##\textrm{sinh u} \frac{du}{dx}=\frac{-1}{x^{2}}##
    ##\frac{du}{dx}=\frac{-1}{x^{2}\textrm{sinh u}}##

    Then I simplify it into
    ##\frac{-1}{x\sqrt{1-x^{2}}}##

    Let ##\frac{dv}{dx}=1##

    ##v=x##

    Using integration by parts and evaluating the integral, I got
    ##\frac{\pi}{2}-\sin^{-1} \frac{4}{5} -\frac{4}{5} \ln 2##

    Which is numerically correct. But how do I obtain ##2\tan^{-1} 2 ## as shown in the question above?
     
  2. jcsd
  3. May 17, 2014 #2

    Curious3141

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    Draw the standard 3-4-5 right triangle. Observe that ##\frac{\pi}{2}-\sin^{-1} \frac{4}{5} = \tan^{-1}\frac{3}{4}##.

    You now have to get ##\tan^{-1}\frac{3}{4}## into something with ##\tan^{-1}2## in it.

    I found this a little tricky. The best solution I could find was to let:

    ##\tan^{-1}\frac{3}{4} = x## so ##\tan x = \frac{3}{4}##

    Then let x = 2y so that ##\tan 2y = \frac{3}{4}##

    Solve for y in the form ##y = \tan^{-1}z##, where z is something you have to find. Only one value is admissible. Express x as 2y = ##2\tan^{-1}z.##

    Now observe that ##\frac{1 + \tan w}{1 - \tan w} = \tan(w + \frac{\pi}{4})##. Use that to find an alternative form for ##\tan^{-1}z##, which will allow you to find x, in the form you need.

    There might be a simpler way (indeed, it might start with an alternative solution of the integral), but I can't immediately find one.
     
    Last edited: May 17, 2014
  4. May 17, 2014 #3

    AlephZero

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    To show the answers are the same, you have to show ##2 \tan^{-1} 2 + \sin^{-1}\displaystyle\frac 4 5 = \pi##.

    From a 3-4-5 triangle, ##\sin^{-1}\displaystyle\frac 4 5 = \tan^{-1}\displaystyle\frac 4 3##.

    From ##\tan^{-1}a + \tan^{-1}b = \tan^{-1}\displaystyle\frac{a+b}{1-ab}##,

    ##2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3##.

    QED.
     
  5. May 22, 2014 #4
    I don't get it, but why
    ##2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3##.
     
  6. May 22, 2014 #5

    CAF123

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    I think the equalities should be $$2 \tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi = \pi - \tan^{-1} \left(\frac{4}{3}\right)$$

    Using Alephzero's formula with ##a=b## gives $$\tan(\tan^{-1} 2 + \tan^{-1} 2) = -\frac{4}{3} \Rightarrow 2\tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi$$
     
  7. May 22, 2014 #6

    SammyS

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    Compare the graphs of ##\ y=\text{arcsech}(x) \ ## and ## y=\text{sech}(x)\ .##

    Integrate ## y=\text{sech}(x)\ ## to get an area related to that given by integrating ##\ y=\text{arcsech}(x) \ .##

    You will have to subtract the area of some rectangle.
     
  8. May 23, 2014 #7
    I plotted two graphs and yet I can't figure it out... :(
    Capture.jpg
     
  9. May 23, 2014 #8

    SammyS

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    The definite integral you are evaluating represents the area below the y = arcsech(x) graph which is between x = 4/5 and x = 1 . Notice that arcsech(4/5) = ln(2) .

    That is the same as the area below the y = sech(x) graph and above y = 4/5, for x ≥ 0. Right?
     

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