TBIL-Calc Tables of Integrals (TI5)

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Section 5.5 Tables of Integrals (TI5)

Learning Outcomes

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I can integrate functions using a table of integrals.

Subsection 5.5.1 Activities

Activity 5.5.1.

Consider the integral

\int \sqrt{16 - 9 x^{2}} d x .

Which of the following substitutions appears most promising to find an antiderivaitve for this integral?

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$u = 16 - 9 x^{2}$
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$u = 9 x^{2}$
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$u = 3 x$
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$u = x$

Activity 5.5.2.

The form of which entry from Appendix A best matches the form of the integral

\int \sqrt{16 - 9 x^{2}} d x ?

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b.
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c.
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g.
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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)

\int \frac{25 x^{2}}{\sqrt{25 x^{2} - 9}} d x

(b)

\int \frac{81 x^{2}}{\sqrt{16 - x^{2}}} d x

(c)

\int \frac{1}{10 x \sqrt{100 - x^{2}}} d x

(d)

\int \frac{7}{\sqrt{25 x^{2} - 9}} d x

(e)

\int \frac{1}{\sqrt{25 x^{2} + 16}} d x

Example 5.5.1.

Here is how one might write out the explanation of how to find

\int \frac{3}{x \sqrt{49 x^{2} - 4}} d x

from start to finish:

\begin{aligned} \int \frac{3}{x \sqrt{49 x^{2} - 4}} d x & Let & u^{2} = 49 x^{2} \\ Let & a^{2} = 4 \\ u = 7 x \\ d u = 7 d x \\ \frac{1}{7} d u = d x \\ a = 2 \\ \int \frac{3}{x \sqrt{49 x^{2} - 4}} d x & = 3 \int \frac{1}{x \sqrt{49 x^{2} - 4}} (d x) \\ = 3 \int \frac{1}{\frac{u}{7} \sqrt{u^{2} - a^{2}}} (\frac{1}{7} d u) \\ = 3 \int \frac{1}{u \sqrt{u^{2} - a^{2}}} d u & which best matches f. \\ = 3 (\frac{1}{a} arcsec (\frac{u}{a})) + C \\ = \frac{3}{2} arcsec (\frac{7 x}{2}) + C \end{aligned}

Activity 5.5.4.

Which step of the previous example do you think was the most important?

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Choosing $u^{2} = 49 x^{2}$ and $a^{2} = 4 .$
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Finding $u = 7 x,$ $d u = 7 d x,$ $\frac{1}{7} d u = d x,$ and $a = 2 .$
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Substituting $\frac{3}{x \sqrt{49 x^{2} - 4}} d x$ with $3 \int \frac{1}{u \sqrt{u^{2} - a^{2}}} d u$ and finding the best match of f from Appendix A.
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Integrating $3 \int \frac{1}{u \sqrt{u^{2} - a^{2}}} d u = 3 (\frac{1}{a} arcsec (\frac{u}{a})) + C .$
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Unsubstituting $3 (\frac{1}{a} arcsec (\frac{u}{a})) + C$ to get $\frac{3}{2} arcsec (\frac{7 x}{2}) + C .$

Activity 5.5.5.

Consider the integral

\int \frac{1}{\sqrt{64 - 9 x^{2}}} d x .

Suppose we proceed using Appendix A. We choose

u^{2} = 9 x^{2}

and

a^{2} = 64 .

(a)

What is

u ?

(b)

What is

d u ?

(c)

What is

a ?

(d)

What do you get when plugging these pieces into the integral

\int \frac{1}{\sqrt{64 - 9 x^{2}}} d x ?

(e)

Is this a good substitution choice or a bad substitution choice?

Activity 5.5.6.

Consider the integral

\int \frac{1}{\sqrt{64 - 9 x^{2}}} d x

once more. Suppose we still proceed using Appendix A. However, this time we choose

u^{2} = x^{2}

and

a^{2} = 64 .

Do you prefer this choice of substitution or the choice we made in Activity 5.5.5?

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We prefer the substitution choice of $u^{2} = x^{2}$ and $a^{2} = 64 .$
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We prefer the substitution choice of $u^{2} = 9 x^{2}$ and $a^{2} = 64 .$
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We do not have a strong preference, since these substitution choices are of the same difficulty.

Activity 5.5.7.

Use the appropriate substitution and entry from Appendix A to show that

\int \frac{7}{x \sqrt{4 + 49 x^{2}}} d x = - \frac{7}{2} \ln | \frac{2 + \sqrt{49 x^{2} + 4}}{7 x} | + C .

Activity 5.5.8.

Use the appropriate substitution and entry from Appendix A to show that

\int \frac{3}{5 x^{2} \sqrt{36 - 49 x^{2}}} d x = - \frac{\sqrt{36 - 49 x^{2}}}{60 x} + C .

Activity 5.5.9.

Evaluate the integral

\int 8 \sqrt{4 x^{2} - 81} d x .

Be sure to specify which entry is used from Appendix A at the corresponding step.

Subsection 5.5.2 Videos

Figure 109. Video: I can integrate functions using a table of integrals

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