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Section 5.4 : More Substitution Rule

10. Evaluate \( \displaystyle \int{{{{\sec }^2}\left( {2t} \right)\left[ {9 + 7\tan \left( {2t} \right) - {{\tan }^2}\left( {2t} \right)} \right]\,dt}}\).

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Hint : Don’t let this one fool you. This is simply an integral that requires you to use the same substitution more than once.
Start Solution

This integral can be a little daunting at first glance. To do it all we need to notice is that the derivative of \(\tan \left( x \right)\) is \({\sec ^2}\left( x \right)\) and we can notice that there is a \({\sec ^2}\left( {2t} \right)\) times the remaining portion of the integrand and that portion only contains constants and tangents.

So, it looks like the substitution is then,

\[u = \tan \left( {2t} \right)\]

If you aren’t comfortable with the basic substitution mechanics you should work some problems in the previous section as we’ll not be putting in as much detail with regards to the basics in this section. The problems in this section are intended for those that are fairly comfortable with the basic mechanics of substitutions and will involve some more “advanced” substitutions.

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Here is the differential work for the substitution.

\[du = 2{\sec ^2}\left( {2t} \right)dt\hspace{0.25in} \to \hspace{0.25in}\,\,\,\,\,{\sec ^2}\left( {2t} \right)dt = \frac{1}{2}du\]

Now, doing the substitution and evaluating the integrals gives,

\[\begin{align*}\int{{{{\sec }^2}\left( {2t} \right)\left[ {9 + 7\tan \left( {2t} \right) - {{\tan }^2}\left( {2t} \right)} \right]\,dt}} & = \frac{1}{2}\int{{9 + 7u - {u^2}du}} = \frac{1}{2}\left( {9u + \frac{7}{2}{u^2} - \frac{1}{3}{u^3}} \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\left( {9\tan \left( {2t} \right) + \frac{7}{2}{{\tan }^2}\left( {2t} \right) - \frac{1}{3}{{\tan }^3}\left( {2t} \right)} \right) + c}}\end{align*}\]

Do not forget to go back to the original variable after evaluating the integral!