*Urgent* trig integral

  • Thread starter frasifrasi
  • Start date
  • #1
276
0
For the integral from sqrt(2) to 2

of
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!
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
Why are there 2 variables in your integral? Is t supposed to be there? Can it be treated as a constant for this question?
 
  • #3
1,753
1
more clarity and a little more work would be appreciated
 
  • #4
276
0
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...
 
  • #5
rock.freak667
Homework Helper
6,223
31
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
 
  • #6
276
0
Oh my god!
 

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