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Section 7.3 : Trig Substitutions

8. Use a trig substitution to eliminate the root in \(\sqrt {{{\bf{e}}^{8x}} - 9} \).

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Hint : This doesn’t look much like a term that can use a trig substitution. So, the first step should probably be to some algebraic manipulation on the quantity under the root to make it look more like a problem that can use a trig substitution.
Start Solution

We know that in order to do a trig substitution we really need a sum or difference of a term with a variable squared and a number. Even though this doesn’t look anything like the “normal” trig substitution problems it is actually pretty close to one. To see this all we need to do is rewrite the term under the root as follows.

\[\sqrt {{{\bf{e}}^{8x}} - 9} = \sqrt {{{\left( {{{\bf{e}}^{4x}}} \right)}^2} - 9} \]

All we did here was take advantage of the basic exponent rules to make it clear that we really do have a difference here of a squared term containing a variable and a number.

Hint : At this point the problem works in the same manner as the previous problems in this section.
Show Step 2

The form of the quantity under the root suggests that secant is the correct trig function to use for the substation.

Now, to get the coefficient on the trig function notice that we need to turn the 1 (i.e. the coefficient of the squared term) into a 9 once we’ve done the substitution. With that in mind it looks like the substitution should be,

\[{{\bf{e}}^{4x}} = 3\sec \left( \theta \right)\]

Now, all we have to do is actually perform the substitution and eliminate the root.

Show Step 3
\[\begin{align*}\sqrt {{{\bf{e}}^{8x}} - 9} & = \sqrt {{{\left( {3\sec \left( \theta \right)} \right)}^2} - 9} = \sqrt {9{{\sec }^2}\left( \theta \right) - 9} \\ & = 3\sqrt {{{\sec }^2}\left( \theta \right) - 1} = 3\sqrt {{{\tan }^2}\left( \theta \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{3\left| {\tan \left( \theta \right)} \right|}}\end{align*}\]

Note that because we don’t know the values of \(\theta \) we can’t determine if the tangent is positive or negative and so cannot get rid of the absolute value bars here.