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Integral of Inverse Cosecant

  1. May 10, 2015 #1
    Running into a little trouble when doing this integral by hand:

    [tex] \int arccsc(x) dx [/tex]

    [tex] u = arccsc(x) \implies du = -\frac{1}{x\sqrt{x^{2}-1}} dx[/tex]

    [tex] dv = dx \implies v = x [/tex]

    [tex] \int u dv = uv - \int vdu [/tex]

    [tex] \int arccsc(x) dx = xarccsc(x) - \int x\cdot -\frac{1}{x\sqrt{x^{2}-1}} dx [/tex]

    [tex] \int arccsc(x) dx = xarccsc(x) + \int \frac{1}{\sqrt{x^{2}-1}} dx [/tex]

    Now I'm here, and it seems obvious , but this yields the answer to be:

    [tex] \int arccsc(x) dx = xarccsc(x) + \ln\left|\sqrt{x^{2}-1}\right| +C [/tex]

    When it's suppose to be (from wikipedia):

    [tex] \int arccsc(x) dx = xarccsc(x) + \ln\left(x+\sqrt{x^{2}-1}\right) +C [/tex]

    Where did I go wrong (skipped ahead in the course, so sorry if it's something really basic I'm missing)?
     
  2. jcsd
  3. May 10, 2015 #2
    That last integral is wrong.
    ##\frac{d}{dx}ln|\sqrt{x^2-1}|=\frac{x}{x^2-1}\neq\frac{1}{\sqrt{x^2-1}}##.
     
    Last edited: May 10, 2015
  4. May 10, 2015 #3
    Yes, I'm aware. I'm wondering where I messed up in the integration by parts.
     
  5. May 10, 2015 #4
    The integration by parts is right.
    However, ##\int \frac{1}{\sqrt{x^2-1}} dx \neq ln|\sqrt{x^2-1}|##.
     
  6. May 10, 2015 #5
    hm..How exactly does ##\int \frac{1}{\sqrt{x^2-1}} dx ## boil down to ##\ln \left( x + \sqrt{x^{2}-1} \right) ## then? I'm a bit lost now.
     
  7. May 10, 2015 #6
    Hint:- Use trigonometric substitution.
     
  8. May 10, 2015 #7
    Thank you! I never heard of trigonometric substitution lol (like I said, I skipped ahead). Did a secant substitution, and got the real answer. Thank you.
     
  9. May 11, 2015 #8
    if you do not want to use trigonometric subst or doing boring calculation, you can multiply and divide ##\left( 1/ \sqrt{x^{2}-1} \right) ## by

    ##\left( x + \sqrt{x^{2}-1} \right) ##

    in this way the result is immediate! In fact you find

    ##\left( 1 + x/\sqrt{x^{2}-1} \right)/ \left( x + \sqrt{x^{2}-1} \right) ##

    that is f'/f , so the result is ##\log \left( x + \sqrt{x^{2}-1} \right) ##


    panta rei
     
    Last edited: May 11, 2015
  10. May 12, 2015 #9
    ..and obviously with the absolute value bars....
     
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