How do you integrate #int 1/sqrt(e^(2x) -9)dx# using trigonometric substitution?

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Answer 1 · 2016-10-01T15:49:43

#1/3arc sec(e^x/3)+C#.

Let #I=int1/sqrt(e^(2x)-9)dx#

We subst. #e^x=3secy," so that, "e^xdx=3sec ytan y dy, i.e., #

#:. 3secydx=3secytanydy, or, dx=tanydy#

Also,

#e^(2x)-9=(3secy)^2-9=9tan^2y rArr sqrt (e^(2x)-9)=3tany.#

Hence, #I=int1/(3tany)tanydy=1/3intdy=1/3y#

Since, #e^x=3secy", we have, "secy=e^x/3, & so, y=arc sec(e^x/3)#.

Finally, #I=1/3arc sec(e^x/3)+C#.