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### Section 7.2 : Integrals Involving Trig Functions

12. Evaluate $$\displaystyle \int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}}$$.

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Hint : How would you do this problem if it were a product?
Start Solution

If this were a product of secants and tangents we would know how to do it. The same ideas work here, except that we have to pay attention to only the numerator. We can’t strip anything out of the denominator (in general) and expect it to work the same way. We can only strip things out of the numerator.

So, let’s notice here is that the exponent on the secant is even and so we can strip out two of them.

$\int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \int{{\frac{{{{\sec }^2}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}{{\sec }^2}\left( {2t} \right)\,dt}}$ Show Step 2

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

$\int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \int{{\frac{{{{\tan }^2}\left( {2t} \right) + 1}}{{{{\tan }^9}\left( {2t} \right)}}{{\sec }^2}\left( {2t} \right)\,dt}}$ Show Step 3

Now we can use the substitution $$u = \tan \left( {2t} \right)$$ to evaluate the integral.

$\int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \frac{1}{2}\int{{\frac{{{u^2} + 1}}{{{u^9}}}\,du}} = \frac{1}{2}\int{{{u^{ - 7}} + {u^{ - 9}}\,du}} = \frac{1}{2}\left[ { - \frac{1}{6}{u^{ - 6}} - \frac{1}{8}{u^{ - 8}}} \right] + c$

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 $$u$$!

$\int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{12}}\frac{1}{{{{\tan }^6}\left( {2t} \right)}} - \frac{1}{{16}}\frac{1}{{{{\tan }^8}\left( {2t} \right)}} + c = - \frac{1}{{12}}{{\cot }^6}\left( {2t} \right) - \frac{1}{{16}}{{\cot }^8}\left( {2t} \right) + c}}$