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Calculus for Team-Based Inquiry Learning 2025 Edition

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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 antiderivative for this integral?
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  1. \(\displaystyle u=16-9x^2\)
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  2. \(\displaystyle u=9x^2\)
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  3. \(\displaystyle u=3x\)
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  4. \(\displaystyle u=x\)
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Activity 5.5.3.

For each of the following integrals, identify which entry from Section A.1 best matches the form of that integral.
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(a)
\(\displaystyle\int \frac{25x^2}{\sqrt{25x^2-9}} \,dx\)
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(b)
\(\displaystyle\int \frac{81x^2}{\sqrt{16-x^2}} \,dx\)
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(c)
\(\displaystyle\int \frac{1}{10x \sqrt{100-x^2}} \,dx\)
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(d)
\(\displaystyle\int \frac{7}{\sqrt{25x^2-9}} \,dx\)
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(e)
\(\displaystyle\int \frac{1}{\sqrt{25x^2+16}} \,dx\)
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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:
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\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*}
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Activity 5.5.5.

Which step of the previous example do you think was the most important?
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  1. Choosing \(u^2=49x^2\) and \(a^2=4\text{.}\)
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  2. Finding \(u=7x\text{,}\) \(du=7\,dx\text{,}\) \(\displaystyle\frac{1}{7}\,du=\,dx\text{,}\) and \(a=2\text{.}\)
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  3. 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 Section A.1.
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  4. Integrating \(\displaystyle 3\int \dfrac{1}{u\sqrt{u^2-a^2}} \,du=3\left(\frac{1}{a}\arcsec\left(\dfrac{u}{a}\right)\right)+C\text{.}\)
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  5. Unsubstituting \(\displaystyle 3\left(\frac{1}{a}\arcsec\left(\frac{u}{a}\right)\right)+C\) to get \(\dfrac{3}{2}\arcsec\left(\dfrac{7x}{2}\right)+C\text{.}\)
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Activity 5.5.6.

Consider the integral \(\displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx\text{.}\) Suppose we proceed using Section A.1. We choose \(u^2=9x^2\) and \(a^2=64\text{.}\)
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(d)
What do you get when plugging these pieces into the integral \(\displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx\text{?}\)
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(e)
Is this a good substitution choice or a bad substitution choice?
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Activity 5.5.7.

Consider the integral \(\displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx\) once more. Suppose we still proceed using Section A.1. 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?
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  1. We prefer the substitution choice of \(u^2=x^2\) and \(a^2=64\text{.}\)
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  2. We prefer the substitution choice of \(u^2=9x^2\) and \(a^2=64\text{.}\)
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  3. We do not have a strong preference, since these substitution choices are of the same difficulty.
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Activity 5.5.8.

Use the appropriate substitution and entry from Section A.1 to show that \(\displaystyle\int \frac{7}{x\sqrt{4+49x^2}} \,dx=-\frac{7}{2}\ln\left|\frac{2+\sqrt{49x^2+4}}{7x}\right|+C\text{.}\)
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Activity 5.5.9.

Use the appropriate substitution and entry from Section A.1 to show that \(\displaystyle\int \frac{3}{5x^2\sqrt{36-49x^2}} \,dx=-\frac{\sqrt{36-49x^2}}{60x}+C\text{.}\)
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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 Section A.1 at the corresponding step.
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Subsection 5.5.2 Videos

Figure 108. Video: I can integrate functions using a table of integrals
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Subsection 5.5.3 Exercises

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