Calculus II - Trig Substitutions

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Section 7.3 : Trig Substitutions

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5. Use a trig substitution to eliminate the root in \(\sqrt {4{{\left( {9t - 5} \right)}^2} + 1} \).

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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 sum of a squared term with a variable in it and a number. This looks similar to the following trig identity (ignoring the coefficients as usual).

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

So, tangent 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.

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Before dealing with the coefficient on the trig function let’s notice that we’ll be substituting in for \(9t - 5\) 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 4 (i.e. the coefficient of the squared term) into a 1 once we’ve done the substitution. With that in mind it looks like the substitution should be,

\[9t - 5 = \frac{1}{2}\tan \left( \theta \right)\]

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

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\[\begin{align*}\sqrt {4{{\left( {9t - 5} \right)}^2} + 1} & = \sqrt {4{{\left( {\frac{1}{2}\tan \left( \theta \right)} \right)}^2} + 1} = \sqrt {4\left( {\frac{1}{4}} \right){{\tan }^2}\left( \theta \right) + 1} = \sqrt {{{\tan }^2}\left( \theta \right) + 1} \\ & = \sqrt {{{\sec }^2}\left( \theta \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{\left| {\sec \left( \theta \right)} \right|}}\end{align*}\]

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.