The integral is: integral(2/(t^2-1),t,2,infinity)
The integral of two over the quantity t squared minus one, from 2 to infinity.
I used trig substitution and ended up in the end with:
2*[sqrt(t^2-1)]+c for the integral, which is not what my calculator gets.
2 times the limit as B goes to infinity of [ln((b^2-1)^(1/2)-ln((3)^(1/2))]
The answer should be ln(3), which is 2*ln((3)^(1/2)), so I thought maybe I'm on the right track, but isn't the limit of as b approaches infinity of
ln((b^2-1)^(1/2), inifinity?
Here are my steps.
t=secu
dt=secutanudu
t^2-1=tan^2u
2*integral((cscdu)/tan^2u)
=2*integral(ln(abs(sinu)/abs(cosu+1)))
sinu= sqrt(t^2-1)/t
cosu=1/t
...
Is there something I am doing wrong.