What is #int 1/(t^2-9)^(1/2)dt#?

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Answer 1 · 2016-07-23T08:51:46

#= ln ( t/3 + sqrt ((t/3)^2 - 1)) + C#

#int 1/(t^2-9)^(1/2)dt#

to use the ID #tan^2 + 1 = sec^2#

let #t^2 = 9 sec^2 theta, t = 3 sec theta, dt = 3 sec theta tan theta d theta qquad triangle#

so the integration becomes

#int 1/(9 sec^2 theta -9)^(1/2) \ 3 sec theta tan theta d theta#

#= int 1/(3 tan theta) \ 3 sec theta tan theta d theta#

#= int sec theta d theta#

which is a standard integral

#= ln ( sec theta + tan theta) + C qquad star#

as in

# int \ d/(d theta) (ln ( sec theta + tan theta) + C) \ d theta#

#= int \ (sec theta tan theta + sec^2 theta)/( sec theta + tan theta) \ d theta#

#= int \ sec theta \ d theta#

reversing the sub in #triangle#, #star# becomes

#= ln ( t/3 + sqrt ((t/3)^2 - 1)) + C#