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### Section 7.3 : Trig Substitutions

2. Use a trig substitution to eliminate the root in $$\sqrt {13 + 25{x^2}}$$.

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Hint : When determining which trig function to use for the substitution recall from the notes in this section that we will use one of three trig identities to convert the sum or difference under the root into a single trig function. Which trig identity is closest to the quantity under the root?
Start Solution

The first step is to figure out which trig function to use for the substitution. To determine this notice that (ignoring the numbers) the quantity under the root looks similar to the identity,

$1 + {\tan ^2}\left( \theta \right) = {\sec ^2}\left( \theta \right)$

So, it looks like tangent is probably the correct trig function to use for the substitution. Now, we need to deal with the numbers on the two terms.

Hint : In order to actually use the identity from the first step we need to get the numbers in each term to be identical upon doing the substitution. So, what would the coefficient of the trig function need to be in order to convert the coefficient of the variable into the constant term once we’ve done the substitution?
Show Step 2

To get the coefficient on the trig function notice that we need to turn the 25 into a 13 once we’ve substituted the trig function in for $$x$$ and squared the substitution out. With that in mind it looks like the substitution should be,

$x = \frac{{\sqrt {13} }}{5}\tan \left( \theta \right)$

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

Show Step 3
\begin{align*}\sqrt {13 + 25{x^2}} & = \sqrt {13 + 25{{\left( {\frac{{\sqrt {13} }}{5}\tan \left( \theta \right)} \right)}^2}} = \sqrt {13 + 25\left( {\frac{{13}}{{25}}} \right){{\tan }^2}\left( \theta \right)} \\ & = \sqrt {13 + 13{{\tan }^2}\left( \theta \right)} = \sqrt {13} \sqrt {1 + {{\tan }^2}\left( \theta \right)} \\ & = \sqrt {13} \sqrt {{{\sec }^2}\left( \theta \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt {13} \left| {\sec \left( \theta \right)} \right|}}\end{align*}

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