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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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 2Now 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.
Show Step 3We 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.
Show Step 4Don’t forget to substitute back in for \(u\)!
\[ \int{{{{\tan }^3}\left( {6x} \right){{\sec }^{10}}\left( {6x} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{72}}{{\sec }^{12}}\left( {6x} \right) - \frac{1}{{60}}{{\sec }^{10}}\left( {6x} \right) + c}}\]