How Do You Evaluate the Integral of arcsec(x) from sqrt(2) to 2?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
frasifrasi
Messages
276
Reaction score
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!
 
Physics news on Phys.org
Why are there 2 variables in your integral? Is t supposed to be there? Can it be treated as a constant for this question?
 
more clarity and a little more work would be appreciated
 
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...
 
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