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Section 7.3 : Trig Substitutions

4. Use a trig substitution to eliminate the root in \(\sqrt {{{\left( {w + 3} \right)}^2} - 100} \).

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Hint : Just because this looks a little different from the first couple of problems in this section doesn’t mean that it works any differently. The term under the root still looks vaguely like one of three trig identities we need to use to convert the quantity under the root into a single trig function.
Start Solution

Okay, first off we need to acknowledge that this does look a little bit different from the first few problems in this section. However, it isn’t really all that different. We still have a difference between a squared term with a variable in it and a number. This looks similar to the following trig identity (ignoring the coefficients as usual).

\[{\sec ^2}\left( \theta \right) - 1 = {\tan ^2}\left( \theta \right)\]

So, secant is the trig function we’ll need to use for the substitution here and we now need to deal with the numbers on the terms and get the substitution set up.

Hint : Dealing with the numbers in this case is no different than the first few problems in this section.
Show Step 2

Before dealing with the coefficient on the trig function let’s notice that we’ll be substituting in for \(w + 3\) in this case since that is the quantity that is being squared in the first term.

So, 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 100 once we’ve done the substitution. With that in mind it looks like the substitution should be,

\[w + 3 = 10\sec \left( \theta \right)\]

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

Show Step 3
\[\begin{align*}\sqrt {{{\left( {w + 3} \right)}^2} - 100} & = \sqrt {{{\left( {10\sec \left( \theta \right)} \right)}^2} - 100} = \sqrt {100{{\sec }^2}\left( \theta \right) - 100} = 10\sqrt {{{\sec }^2}\left( \theta \right) - 1} \\ & = 10\sqrt {{{\tan }^2}\left( \theta \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{10\left| {\tan \left( \theta \right)} \right|}}\end{align*}\]

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.