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Section 7.2 : Integrals Involving Trig Functions

6. Evaluate \( \displaystyle \int{{{{\tan }^3}\left( {6x} \right){{\sec }^{10}}\left( {6x} \right)\,dx}}\).

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Hint : Pay attention to the exponents and recall that for most of these kinds of problems you’ll need to use trig identities to put the integral into a form that allows you to do the integral (usually with a Calc I substitution).
Start Solution

The first thing to notice here is that the exponent on the tangent is odd and we’ve got a secant in the problems and so we can strip one of each of them out.

\[ \int{{{{\tan }^3}\left( {6x} \right){{\sec }^{10}}\left( {6x} \right)\,dx}} = \int{{{{\tan }^2}\left( {6x} \right){{\sec }^9}\left( {6x} \right)\,\,\,\tan \left( {6x} \right)\sec \left( {6x} \right)\,dx}}\] Show Step 2

Now we can use the trig identity \({\tan ^2}\theta + 1 = {\sec ^2}\theta \) to convert the remaining tangents to secants.

\[ \int{{{{\tan }^3}\left( {6x} \right){{\sec }^{10}}\left( {6x} \right)\,dx}} = \int{{\left[ {{{\sec }^2}\left( {6x} \right) - 1} \right]{{\sec }^9}\left( {6x} \right)\,\,\,\tan \left( {6x} \right)\sec \left( {6x} \right)\,dx}}\]

Note that because the exponent on the secant is even we could also have just stripped two of them out and converted the rest of them to tangents. However, that conversion process would have been significantly more work than the path that we chose here.

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We can now use the substitution \(u = \sec \left( {6x} \right)\) to evaluate the integral.

\[\begin{align*}\int{{{{\tan }^3}\left( {6x} \right){{\sec }^{10}}\left( {6x} \right)\,dx}} & = \frac{1}{6}\int{{\left[ {{u^2} - 1} \right]{u^9}\,du}}\\ & = \frac{1}{6}\int{{{u^{11}} - {u^9}\,du}} = \frac{1}{6}\left( {\frac{1}{{12}}{u^{12}} - \frac{1}{{10}}{u^{10}}} \right) + c\end{align*}\]

Note that we’ll not be doing the actual substitution work here. At this point it is assumed that you recall substitution well enough to fill in the details if you need to. If you are rusty on substitutions you should probably go back to the Calculus I practice problems and practice on the substitutions.

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Don’t forget to substitute back in for \(u\)!

\[ \int{{{{\tan }^3}\left( {6x} \right){{\sec }^{10}}\left( {6x} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{72}}{{\sec }^{12}}\left( {6x} \right) - \frac{1}{{60}}{{\sec }^{10}}\left( {6x} \right) + c}}\]