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

\[\begin{align*}\int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}} & = \int{{{{\left[ {{{\sec }^2}\left( {3y} \right)} \right]}^2}{{\tan }^2}\left( {3y} \right){{\sec }^2}\left( {3y} \right)\,dy}}\\ & = \int{{{{\left[ {{{\tan }^2}\left( {3y} \right) + 1} \right]}^2}{{\tan }^2}\left( {3y} \right){{\sec }^2}\left( {3y} \right)\,dy}}\end{align*}\] Show Step 3

We can now use the substitution \(u = \tan \left( {3y} \right)\) to evaluate the integral.

\[\begin{align*}\int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}}& = \frac{1}{3}\int{{{{\left[ {{u^2} + 1} \right]}^2}{u^2}\,du}}\\ & = \frac{1}{3}\int{{{u^6} + 2{u^4} + {u^2}\,du}} = \frac{1}{3}\left( {\frac{1}{7}{u^7} + \frac{2}{5}{u^5} + \frac{1}{3}{u^3}} \right) + c\end{align*}\]

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 4

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

\[\int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{21}}{{\tan }^7}\left( {3y} \right) + \frac{2}{{15}}{{\tan }^5}\left( {3y} \right) + \frac{1}{9}{{\tan }^3}\left( {3y} \right) + c}}\]