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Section 7.2 : Integrals Involving Trig Functions

7. Evaluate \( \displaystyle \int_{0}^{{\frac{\pi }{4}}}{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}}\).

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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

We have two options for dealing with the limits. We can deal with the limits as we do the integral or we can evaluate the indefinite integral and take care of the limits in the last step. We’ll use the latter method of dealing with the limits for this problem.

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

\[\begin{align*}\int{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} & = \int{{{{\tan }^6}\left( z \right){{\sec }^2}\left( z \right)\,\,\,\tan \left( z \right)\sec \left( z \right)\,dz}}\\ & = \int{{{{\left[ {{{\tan }^2}\left( z \right)} \right]}^3}{{\sec }^2}\left( z \right)\,\,\,\tan \left( z \right)\sec \left( z \right)\,dz}}\\ & = \int{{{{\left[ {{{\sec }^2}\left( z \right) - 1} \right]}^3}{{\sec }^2}\left( z \right)\,\,\,\tan \left( z \right)\sec \left( z \right)\,dz}}\end{align*}\] Show Step 2

We can now use the substitution \(u = \sec \left( z \right)\) to evaluate the integral.

\[\begin{align*}\int{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} & = \int{{{{\left[ {{u^2} - 1} \right]}^3}{u^2}\,du}}\\ & = \int{{{u^8} - 3{u^6} + 3{u^4} - {u^2}\,du}} = \frac{1}{9}{u^9} - \frac{3}{7}{u^7} + \frac{3}{5}{u^5} - \frac{1}{3}{u^3} + c\end{align*}\]

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 \(u\)!

\[\int{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} = \frac{1}{9}{\sec ^9}\left( z \right) - \frac{3}{7}{\sec ^7}\left( z \right) + \frac{3}{5}{\sec ^5}\left( z \right) - \frac{1}{3}{\sec ^3}\left( z \right) + c\] Show Step 4

Now all we need to do is deal with the limits.

\[\begin{align*}\int_{0}^{{\frac{\pi }{4}}}{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} & = \left. {\left( {\frac{1}{9}{{\sec }^9}\left( z \right) - \frac{3}{7}{{\sec }^7}\left( z \right) + \frac{3}{5}{{\sec }^5}\left( z \right) - \frac{1}{3}{{\sec }^3}\left( z \right)} \right)} \right|_0^{\frac{\pi }{4}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{{315}}\left( {8 + 13\sqrt 2 } \right) = 0.1675}}\end{align*}\]