Calculus II - Trig Substitutions

Show Mobile Notice Show All Notes Hide All Notes

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7.3 : Trig Substitutions

Back to Problem List

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

Show All Steps Hide All Steps

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.

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.