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### Section 7.3 : Trig Substitutions

13. Use a trig substitution to evaluate $$\displaystyle \int_{1}^{4}{{2{z^5}\sqrt {2 + 9{z^2}} \,dz}}$$ .

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Start Solution

In this case it looks like we’ll need the following trig substitution.

$z = \frac{{\sqrt 2 }}{3}\tan \left( \theta \right)$

Now we need to use the substitution to eliminate the root and get set up for actually substituting this into the integral.

Show Step 2

Let’s first use the substitution to eliminate the root.

$\sqrt {2 + 9{z^2}} = \sqrt {2 + 2{{\tan }^2}\left( \theta \right)} = \sqrt 2 \sqrt {{{\sec }^2}\left( \theta \right)} = \sqrt 2 \left| {\sec \left( \theta \right)} \right|$ Show Step 3

Okay, in this case we have limits on $$z$$ and so we can get limits on $$\theta$$ that will allow us to determine if tangent is positive or negative to allow us to eliminate the absolute value bars.

So, let’s get some limits on $$\theta$$.

\begin{align*}z & = 1:\,\,\,\,\,1 = \frac{{\sqrt 2 }}{3}\tan \left( \theta \right)\hspace{0.25in} \to \hspace{0.25in}\tan \left( \theta \right) = \frac{3}{{\sqrt 2 }}\hspace{0.25in} \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 2 }}} \right) = 1.1303\\ z & = 4:\,\,\,\,4 = \frac{{\sqrt 2 }}{3}\tan \left( \theta \right)\hspace{0.25in} \to \hspace{0.25in}\tan \left( \theta \right) = \frac{{12}}{{\sqrt 2 }}\hspace{0.25in} \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( {\frac{{12}}{{\sqrt 2 }}} \right) = 1.4535\end{align*}

So, $$\theta$$’s for this problem are in the range $$1.1303 \le \theta \le 1.4535$$ and these are in the first quadrant. In the first quadrant we know that cosine, and hence secant, is positive and so we can just drop the absolute value bars. This gives,

$\sqrt {2 + 9{z^2}} = \sqrt 2 \sec \left( \theta \right)$

For a final substitution preparation step let’s also compute the differential so we don’t forget to use that in the substitution!

$dz = \frac{{\sqrt 2 }}{3}{\sec ^2}\left( \theta \right)\,d\theta$ Show Step 4

Now let’s do the actual substitution.

\begin{align*}\int_{1}^{4}{{2{z^5}\sqrt {2 + 9{z^2}} \,dz}} & = \int_{{1.1303}}^{{1.4535}}{{2{{\left( {\frac{{\sqrt 2 }}{3}} \right)}^5}{{\tan }^5}\left( \theta \right)\left( {\sqrt 2 \sec \left( \theta \right)} \right)\,\left( {\frac{{\sqrt 2 }}{3}{{\sec }^2}\left( \theta \right)} \right)d\theta }}\\ & = \frac{{16\sqrt 2 }}{{729}}\int_{{1.1303}}^{{1.4535}}{{{{\tan }^5}\left( \theta \right){{\sec }^3}\left( \theta \right)d\theta }}\end{align*}

Do not forget to substitute in the differential we computed in the previous step. This is probably the most common mistake with trig substitutions. Forgetting the differential can substantially change the problem, often making the integral very difficult to evaluate.

Also notice that upon doing the substation we replaced the $$y$$ limits with the $$\theta$$ limits. This will help with a later step.

Show Step 5

We now need to evaluate the integral. Here is that work.

\begin{align*}\int_{1}^{4}{{2{z^5}\sqrt {2 + 9{z^2}} \,dz}} & = \frac{{16\sqrt 2 }}{{729}}\int_{{1.1303}}^{{1.4535}}{{{{\left[ {{{\sec }^2}\left( \theta \right) - 1} \right]}^2}{{\sec }^2}\left( \theta \right)\,\,\tan \left( \theta \right)\sec \left( \theta \right)d\theta }}\\ & = \frac{{16\sqrt 2 }}{{729}}\int_{{\sec \left( {1.1303} \right)}}^{{\sec \left( {1.4535} \right)}}{{{{\left[ {{u^2} - 1} \right]}^2}{u^2}du}}\hspace{0.75in}u = \sec \left( \theta \right)\\ & = \frac{{16\sqrt 2 }}{{729}}\int_{{2.3452}}^{{8.5440}}{{{u^6} - 2{u^4} + {u^2}du}}\\ & = \left. {\frac{{16\sqrt 2 }}{{729}}\left[ {\frac{1}{7}{u^7} - \frac{2}{5}{u^5} + \frac{1}{3}{u^3}} \right]} \right|_{2.3452}^{8.5440} = \require{bbox} \bbox[2pt,border:1px solid black]{{14178.20559}}\end{align*}

Don’t forget all the “standard” manipulations of the integrand that we often need to do in order to evaluate integrals involving trig functions. If you don’t recall them you’ll need to go back to the previous section and work some practice problems to get good at them.

Every trig substitution problem reduces down to an integral involving trig functions and the majority of them will need some manipulation of the integrand in order to evaluate.

Also, note that because we converted the limits at every substitution into limits for the “new” variable we did not need to do any back substitution work on our answer!