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Homework Help: *Urgent* trig integral

  1. Dec 16, 2007 #1
    For the integral from sqrt(2) to 2

    1 over x*sqrt(t^(2) - 1) dx

    I noticed that this was just the arcsece, so I got arcsec(x) for the answer, but how would I evaluated this at 2 and sqrt(2)?????

    What did i do wrong?

    Thank you!
  2. jcsd
  3. Dec 16, 2007 #2


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    Why are there 2 variables in your integral? Is t supposed to be there? Can it be treated as a constant for this question?
  4. Dec 16, 2007 #3
    more clarity and a little more work would be appreciated
  5. Dec 16, 2007 #4
    Ok, the integral is:

    1 over x*sqrt(x^(2) - 1) dx

    --> which I evaluated to be arcsec (x),but this doesn't make sense with the limits of integration...
  6. Dec 16, 2007 #5


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    Well if you want to find arcsec([itex]\sqrt{2}[/itex]) you can always work it out like this:

    Let [itex]\alpha=sec^{-1}(\sqrt{2})[/itex]

    so that [itex]sec\alpha=\sqrt{2}[/itex]
    and therefore [itex]cos\alpha=\frac{1}{\sqrt{2}}[/itex] and then you find [itex]\alpha[/itex]

    OR...somewhere in you attempt you would have used the substitution x=sec[itex]\theta[/itex] so from there you could have gotten [itex]\theta=cos^{-1}(\frac{1}{x})[/itex] and use that instead of arcsec
  7. Dec 16, 2007 #6
    Oh my god!
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