Learning Outcomes
- I can integrate functions using a table of integrals.
Section 5.5 Tables of Integrals (TI5)
Subsection 5.5.1 Activities
Activity 5.5.1.
Consider the integral \(\displaystyle\int \sqrt{16-9x^2} \,dx\text{.}\) Which of the following substitutions appears most promising to find an antiderivaitve for this integral?
- \(\displaystyle u=16-9x^2\)
- \(\displaystyle u=9x^2\)
- \(\displaystyle u=3x\)
- \(\displaystyle u=x\)
Activity 5.5.2.
The form of which entry from Appendix A best matches the form of the integral \(\displaystyle\int \sqrt{16-9x^2} \,dx\text{?}\)
- b.
- c.
- g.
- h.
Activity 5.5.3.
For each of the following integrals, identify which entry from Appendix A best matches the form of that integral.
(a)
\(\displaystyle\int \frac{25x^2}{\sqrt{25x^2-9}} \,dx\)
(b)
\(\displaystyle\int \frac{81x^2}{\sqrt{16-x^2}} \,dx\)
(c)
\(\displaystyle\int \frac{1}{10x \sqrt{100-x^2}} \,dx\)
(d)
\(\displaystyle\int \frac{7}{\sqrt{25x^2-9}} \,dx\)
(e)
\(\displaystyle\int \frac{1}{\sqrt{25x^2+16}} \,dx\)
Example 5.5.4.
Here is how one might write out the explanation of how to find \(\displaystyle\int \frac{3}{x\sqrt{49x^2-4}} \,dx\) from start to finish:
\begin{align*}
\int \frac{3}{x\sqrt{49x^2-4}} \,dx &&\text{Let }&u^2=49x^2\\
&&\text{Let }&a^2=4 \\
&&& u = 7x\\
&&& \,du = 7\,dx\\
&&& \frac{1}{7}\,du = \,dx\\
&&& a = 2\\
\int \frac{3}{x\sqrt{49x^2-4}} \,dx &= 3\int \frac{1}{x\sqrt{49x^2-4}} (\,dx)\\
&= 3\int \frac{1}{\frac{u}{7}\sqrt{u^2-a^2}} \bigg(\frac{1}{7}\,du\bigg)\\
&= 3\int \frac{1}{u\sqrt{u^2-a^2}} \,du & \text{which best matches f.}\\
&= 3\bigg(\frac{1}{a}\arcsec \bigg(\frac{u}{a}\bigg)\bigg)+C\\
&= \frac{3}{2}\arcsec \bigg(\frac{7x}{2}\bigg)+C
\end{align*}
Activity 5.5.5.
Which step of the previous example do you think was the most important?
- Choosing \(u^2=49x^2\) and \(a^2=4\text{.}\)
- Finding \(u=7x\text{,}\) \(du=7\,dx\text{,}\) \(\displaystyle\frac{1}{7}\,du=\,dx\text{,}\) and \(a=2\text{.}\)
- Substituting \(\displaystyle \frac{3}{x\sqrt{49x^2-4}} \,dx\) with \(\displaystyle3\int \frac{1}{u\sqrt{u^2-a^2}} \,du\) and finding the best match of f from Appendix A.
- Integrating \(\displaystyle3\int \frac{1}{u\sqrt{u^2-a^2}} \,du=3(\frac{1}{a}\arcsec(\frac{u}{a}))+C\text{.}\)
- Unsubstituting \(\displaystyle3(\frac{1}{a}\arcsec(\frac{u}{a}))+C\) to get \(\displaystyle\frac{3}{2}\arcsec(\frac{7x}{2})+C\text{.}\)
Activity 5.5.6.
Consider the integral \(\displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx\text{.}\) Suppose we proceed using Appendix A. We choose \(u^2=9x^2\) and \(a^2=64\text{.}\)
(a)
What is \(u\text{?}\)
(b)
What is \(du\text{?}\)
(c)
What is \(a\text{?}\)
(d)
What do you get when plugging these pieces into the integral \(\displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx\text{?}\)
(e)
Is this a good substitution choice or a bad substitution choice?
Activity 5.5.7.
Consider the integral \(\displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx\) once more. Suppose we still proceed using Appendix A. However, this time we choose \(u^2=x^2\) and \(a^2=64\text{.}\) Do you prefer this choice of substitution or the choice we made in Activity 5.5.6?
- We prefer the substitution choice of \(u^2=x^2\) and \(a^2=64\text{.}\)
- We prefer the substitution choice of \(u^2=9x^2\) and \(a^2=64\text{.}\)
- We do not have a strong preference, since these substitution choices are of the same difficulty.
Activity 5.5.8.
Use the appropriate substitution and entry from Appendix A to show that \(\displaystyle\int \frac{7}{x\sqrt{4+49x^2}} \,dx=-\frac{7}{2}\ln\big|\frac{2+\sqrt{49x^2+4}}{7x}\big|+C\text{.}\)
Activity 5.5.9.
Use the appropriate substitution and entry from Appendix A to show that \(\displaystyle\int \frac{3}{5x^2\sqrt{36-49x^2}} \,dx=-\frac{\sqrt{36-49x^2}}{60x}+C\text{.}\)
Activity 5.5.10.
Evaluate the integral \(\displaystyle\int 8\sqrt{4x^2-81} \,dx\text{.}\) Be sure to specify which entry is used from Appendix A at the corresponding step.
