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Basic indefinite integral

  1. Nov 16, 2011 #1
    Hi guys, I don't even know where to begin with this question.

    Find the following indefinite integral:

    [tex] \int x.arsinh (x^2) dx [/tex]

    Thanks very much for any help, its much appreciated.
     
  2. jcsd
  3. Nov 16, 2011 #2

    Char. Limit

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    Gold Member

    Well, first you'll need a u-substitution and then integration by parts. The u-sub is simple, so I'll skip that part. For the integration by parts, try putting u=arcsin(x) and dv=dx.
     
  4. Nov 16, 2011 #3
    Thanks for that, I got the following final answer:

    [tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

    Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?
     
    Last edited: Nov 16, 2011
  5. Nov 16, 2011 #4
  6. Nov 16, 2011 #5

    Mark44

    Staff: Mentor

    The substitution should be u = arcsinh(x). There is no arsinh function.
     
  7. Nov 16, 2011 #6

    Char. Limit

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    Gold Member

    Woops! I thought that read arcsin(x)! My bad.
     
  8. Nov 16, 2011 #7
    Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?
     
  9. Nov 16, 2011 #8

    Mark44

    Staff: Mentor

    Did your teacher also write arcsin and arccos as arsin and arcos?
     
  10. Nov 16, 2011 #9
    No they were written as arcsin and arccos. It was just the hyperbolic ones that were written without a c.
     
  11. Nov 16, 2011 #10

    Char. Limit

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    arsinh and arcosh are the correct forms; they stand for "area hyperbolic sine" and "area hyperbolic cosine", respectively.
     
  12. Nov 16, 2011 #11

    Mark44

    Staff: Mentor

    Well, ya learn something every day.
     
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