How do you integrate $\int \frac{1}{\sqrt{e^{2 x} - 9}} d x$ $int 1/sqrt(e^(2x) -9)dx$ using trigonometric substitution?

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Answer 1 · 2016-10-01T15:49:43

$\frac{1}{3} a r c sec (\frac{e^{x}}{3}) + C$ $1/3arc sec(e^x/3)+C$ .

Let $I = \int \frac{1}{\sqrt{e^{2 x} - 9}} d x$ $I=int1/sqrt(e^(2x)-9)dx$

We subst. $e^{x} = 3 sec y, so that, e^{x} d x = 3 sec y tan y d y, i . e .,$ $e^x=3secy," so that, "e^xdx=3sec ytan y dy, i.e.,$

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

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

How do you integrate int 1/sqrt(e^(2x) -9)dx∫1√e2x−9dxint 1/sqrt(e^(2x) -9)dx using trigonometric substitution?