Question

How do you find the integral of `int 1/sqrt(t^2-36) dt` from 6 to infinity?

Answer 1

The integral diverges.

Explanation:

`int_6^oo 1/sqrt(t^2-36) dt`

The integrand is not defined at `t=6`, so we need to split the integral and use two limits. I choose to split the integral at `7` (any number greater than `6` will work).

`= int_6^7 1/sqrt(t^2-36) dt+int_7^oo 1/sqrt(t^2-36) dt` `" "` (If the integrals both exist.)

`= lim_(ararr6^+) int_a^7 1/sqrt(t^2-36) dt+lim_(brarroo)int_7^b 1/sqrt(t^2-36) dt` `" "` (If the limits both exist.)

Now we need the indefinite integral

`int 1/sqrt(t^2-36) dt`

Substitute `t = secu`, so that `dt = 6secutanu du` and the integral (after substitution) simplifies to

`int secu du`

Which, if you don't know it, can be found here at Socratic or in most calculus textbooks, etc.

`int secu du = ln(tanu+secu)+C`

Back-substituting for `u` gets us:

`ln(1/6(sqrt(t^2-36)+t))+C`, or, more cleanly:

`ln(sqrt(t^2-36)+t)+C`,

`int_6^oo 1/sqrt(t^2-36) dt`

`= lim_(ararr6^+) [ln(sqrt(7^2-36)+7)-ln(sqrt(a^2-36)+a)]`

`+ lim_(brarroo)[ln(sqrt(b^2-36)+b)-ln(sqrt(7^2-36)+7)]`

(If the limits both exist.)

The second limit diverges, so the integral diverges.