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Trigonometric Integral Confusion

  1. Jul 12, 2010 #1

    Char. Limit

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

    So, for absolutely no good reason I decided to practice trigonometric substitution. I started with [itex]\int \left(1-x^2\right)^{-1/2} dx[/itex], and that was easy, everything cancelled out nicely. Then I tried [itex]\int \left(1+x^2\right)^{-1/2} dx[/itex], and although the integral was fine, everything did not cancel out nicely afterward. So I need help on one minor issue.

    2. Relevant equations

    [tex] sec(u) = \frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)}[/tex]

    3. The attempt at a solution

    [tex]\int \frac{dx}{\sqrt{1+x^2}}[/tex]

    [tex]\int \frac{sec^2(u) du}{\sqrt{1+tan^2(u)}}[/tex]

    [tex]\int sec(u) du[/tex]

    [tex]\int \frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)} du[/tex]

    [tex]\int \frac{dv}{v}[/tex]

    [tex]ln(v)[/tex]

    [tex]ln(tan(u) + sec(u))[/tex]

    [tex]ln(x + sec(tan^{-1}(x)))[/tex]



    Now, is there any way to simplify sec(arctan(x))?
     
  2. jcsd
  3. Jul 12, 2010 #2
    Let θ = arctan(x), then tan(θ) = x = x/1. Draw a right triangle, label the sides, and then find sec(θ).
     
  4. Jul 12, 2010 #3

    vela

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    Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1.
     
  5. Jul 12, 2010 #4

    Char. Limit

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    Hmm... I did that, but I don't think my answer is right... the O=x, A=1, H=sqrt(1+x^2) idea gives me sec(arctan(x))=sqrt(1+x^2), and that just doesn't sound right... but it probably is.

    So...

    [tex]ln\left(x+\sqrt{1+x^2}\right) +C[/tex]

    Is the answer? For some reason I expected a trig function to pop up like it did with 1/sqrt(1-x^2)...
     
  6. Jul 12, 2010 #5
    That's the right answer, you can always check on Wolfram|Alpha too.
     
  7. Jul 12, 2010 #6

    vela

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    Or you could differentiate your answer and see if you get the integrand back.

    If you think in terms of the hyperbolic trig functions, your answer is arcsinh(x), so a trig function does actually pop up.
     
  8. Jul 12, 2010 #7

    Char. Limit

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    Oh I see. Thanks for the info on the arsinh(x) function... I just looked it up on Wikipedia, and it helped me understand that there was a trig function hidden in that logarithmic function.
     
  9. Jul 13, 2010 #8

    hunt_mat

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    There are two ways to solve this integral. There is the use of [tex]x=\tan\theta[/tex] and by the use of the identity [tex]1+\tan^{2}theta =\sec^{2}\theta[/tex] and the use of [tex]x=\sinh u[/tex] and use [tex]\cosh^{2}u=1+\sinh^{2}u[/tex]. I personally think it is easier with the hyperbolic function.
     
  10. Jul 13, 2010 #9

    Char. Limit

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    Yes, but I had never heard of the hyperbolic functions before today, so I don't quite know the derivative of sinh and cosh. Is it the same as the derivative of sin and cos? Namely...

    [tex]\frac{d sinh(x)}{dx} = cosh(x)[/tex]

    [tex]\frac{d cosh(x)}{dx} = - sinh(x)[/tex]

    Are those correct?
     
  11. Jul 13, 2010 #10

    hunt_mat

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    No. The definitions are as follows:
    [tex]
    \sinh x=\frac{e^{x}-e^{-x}}{2}\quad\cosh x=\frac{e^{x}+e^{-x}}{2}
    [/tex]
    From these definitions, you can calculate the derivatives and also prove the basic identity:
    [tex]
    \cosh^{2}x-\sinh^{2}x=1
    [/tex]
    From these three definitions, you can pretty much prove everything you need.

    Mat
     
  12. Jul 13, 2010 #11

    Char. Limit

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    Oh, I see. So the derivative is a two-point cycle, not a four-point cycle. Thank you.
     
  13. Jul 13, 2010 #12

    hunt_mat

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    There's not much to learn about hyperbolic functions apart from their definitions, their graphs and their derivatives. I am not sure of their geometric interpretation, there must be one, hyperbolic geometry perhaps?

    mat
     
  14. Jul 13, 2010 #13

    Char. Limit

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    I just looked it up, they are supposed to give measures of a hyperbola, I believe. Somehow.
     
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