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

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

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