###
Subsection 5.12.1 Using an Integral Table

Calculus has a rich history, spanning across many centuries and continents. Core concepts like limit and continuity can be traced back to Greek mathematicians, astronomers, and philosophers in 400-300 BC. Scholars in China, India, the Middle East, and Europe developed these ideas further over hundreds of years until the first unified accounts of differential and integral calculus were given, independently, by Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century.

It is instructive to realize that because the personal computer didn’t exist until the late 1980s, calculus (and all other mathematics) had to be done by hand for millenia. In the 21st century, however, computers have revolutionized many aspects of the world we live in, including mathematics. In this section we investigate the role that integral tables and computer algebra systems can play in evaluating indefinite integrals.

As seen in the short table of integrals found in

Appendix A, there are many forms of integrals that involve

\(\sqrt{a^2 \pm w^2}\) and

\(\sqrt{w^2 - a^2}\text{.}\) These integral rules can be developed using the technique known as

*trigonometric substitution* that we learned in the last section. To see how these rules are used, consider the differences among

\begin{equation*}
\int \frac{1}{\sqrt{1-x^2}} \,dx, \ \ \ \int \frac{x}{\sqrt{1-x^2}} \,dx, \ \ \ \text{and} \ \ \ \int \sqrt{1-x^2} \,dx\text{.}
\end{equation*}

The first integral is a familiar basic one, and results in \(\arcsin(x) + C\text{.}\) The second integral can be evaluated using a standard \(u\)-substitution with \(u = 1-x^2\text{.}\) The third, however, is not familiar and does not lend itself to \(u\)-substitution.

\begin{equation*}
\text{(h)} ~ \int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2} \arcsin \frac{u}{a} + C\text{.}
\end{equation*}

Using the substitutions \(a = 1\) and \(u = x\) (so that \(du = dx\)), it follows that

\begin{equation*}
\int \sqrt{1-x^2} \, dx = \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \arcsin x + C\text{.}
\end{equation*}

Whenever we are applying a rule in the table, we are doing a \(u\)-substitution, particularly when the substitution is more complicated than setting \(u = x\) as in the last example.

####
Example 5.110.

Evaluate the integral

\begin{equation*}
\int \sqrt{9 + 64x^2} \, dx\text{.}
\end{equation*}

## Solution.

Here, we want to use Rule (c) from the table, but we now set \(a = 3\) and \(u = 8x\text{.}\) We also choose the “\(+\)” option in the rule. With this substitution, it follows that \(du = 8dx\text{,}\) so \(dx = \frac{1}{8} du\text{.}\) Applying the substitution,

\begin{equation*}
\int \sqrt{9 + 64x^2} \, dx = \int \sqrt{9 + u^2} \cdot \frac{1}{8} \, du = \frac{1}{8} \int \sqrt{9+u^2} \, du\text{.}
\end{equation*}

By Rule (c), we now find that

\begin{align*}
\int \sqrt{9 + 64x^2} \, dx =\mathstrut \amp \frac{1}{8} \left( \frac{u}{2}\sqrt{u^2 + 9} + \frac{9}{2}\ln\left|u + \sqrt{u^2 + 9}\right| + C \right)\\
=\mathstrut \amp \frac{1}{8} \left( \frac{8x}{2}\sqrt{64x^2 + 9} + \frac{9}{2}\ln\left|8x + \sqrt{64x^2 + 9}\right| + C \right)\text{.}
\end{align*}

Whenever we use a

\(u\)-subsitution in conjunction with

Appendix A, it’s important that we not forget to address any constants that arise and include them in our computations, such as the

\(\frac{1}{8}\) that appeared in

Example 5.110.

####
Example 5.111.

For each of the following integrals, evaluate the integral using

\(u\)-substitution and/or an entry from the table found in

Appendix A.

\(\displaystyle \int \sqrt{x^2 + 4} \, dx\)

\(\displaystyle \int \frac{x}{\sqrt{x^2 +4}} \, dx\)

\(\displaystyle \int \frac{2}{\sqrt{16+25x^2}}\, dx\)

\(\displaystyle \int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx\)

## Hint.

Compare to \(\int \sqrt{u^2 + a^2} \, du\text{.}\)

Try a straightforward \(u\)-substitution; the table is unneeded.

Let \(a = 4\) and \(u = 5x\) and look for a similar integral in the table.

Let \(a = 7\) and \(u = 6x\text{;}\) find a related integral in the table.

## Answer.

\(\int \sqrt{x^2 + 4} \, dx = \frac{x}2\sqrt{x^2+4}+2\ln\big|x+\sqrt{x^2+4}\big|+C\text{.}\)

\(\int \frac{x}{\sqrt{x^2 +4}} \, dx = \sqrt{x^2 + 4} + C\text{.}\)

\(\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln\big| 5x + \sqrt{16+25x^2} \big| + C\text{.}\)

\(\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = - \frac{\sqrt{49-36x^2}}{49x} + C\text{.}\)

## Solution.

By Rule (c) in

Appendix A with

\(a=2\) and

\(u = x\text{,}\)
\begin{equation*}
\int \sqrt{x^2 + 4} \, dx = \frac{x}2\sqrt{x^2+4}+\frac42\ln\left|x+\sqrt{x^2+4}\right|+C\text{.}
\end{equation*}

Let \(u=x^2 + 4\text{,}\) so \(du = 2x\, dx\text{.}\) Thus

\begin{equation*}
\int \frac{x}{\sqrt{x^2 +4}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{x^2 + 4} + C\text{.}
\end{equation*}

Letting \(a = 4\) and \(u = 5x\text{,}\) we see \(du=5\,dx\text{.}\) Thus,

\begin{equation*}
\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \int \frac{du}{\sqrt{a^2 + u^2}}\text{.}
\end{equation*}

\begin{equation*}
\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln\left| 5x + \sqrt{16+25x^2} \right| + C\text{.}
\end{equation*}

Letting \(a = 7\) and \(u = 6x\text{,}\) it follows that \(x=\frac{1}{6}u\) and \(du=6\,dx\text{,}\) and therefore

\begin{equation*}
\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = \frac{1}{6} \int \frac{1}{\frac{1}{36}u^2 \sqrt{a^2-u^2}} \, du\text{.}
\end{equation*}

Using Rule (k) in

Appendix A and the fact that

\(\frac{1}{6} \cdot 36 = 6\text{,}\) we see

\begin{equation*}
\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = -6\cdot \frac{\sqrt{49-36x^2}}{49 \cdot 6x} + C\text{.}
\end{equation*}

###
Subsection 5.12.2 Using Computer Algebra Systems

A computer algebra system (CAS) is a computer program that is capable of executing symbolic mathematics. For example, if we ask a CAS to solve the equation \(ax^2 + bx + c = 0\) for the variable \(x\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are arbitrary constants, the program will return \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.}\) Research to develop the first CAS dates to the 1960s, and these programs became publicly available in the early 1990s. Two prominent examples are the programs *Maple* and *Mathematica*, which were among the first computer algebra systems to offer a graphical user interface. Today, *Maple* and *Mathematica* are exceptionally powerful professional software packages that can execute an amazing array of sophisticated mathematical computations. They are also very expensive, as each is a proprietary program. The CAS *SAGE* is an open-source, free alternative to *Maple* and *Mathematica*.

For the purposes of this text, when we need to use a CAS, we are going to turn instead to a similar, but somewhat different computational tool, the web-based “computational knowledge engine” called *WolframAlpha*. There are two features of *WolframAlpha* that make it stand out from the CAS options mentioned above: (1) unlike *Maple* and *Mathematica*, *WolframAlpha* is free (provided we are willing to navigate some pop-up advertising); and (2) unlike any of the three, the syntax in *WolframAlpha* is flexible. Think of *WolframAlpha* as being a little bit like doing a Google search: the program will interpret what is input, and then provide a summary of options.

If we want to have *WolframAlpha* evaluate an integral for us, we can provide it syntax such as

`integrate x^2 dx`

to which the program responds with

\begin{equation*}
\int x^2 \, dx = \frac{x^3}{3} + \text{constant}\text{.}
\end{equation*}

To find the partial fraction decomposition of any rational function, in *WolframAlpha*, entering

`partial fraction 5x/(x^2-x-2)`

results in the output

\begin{equation*}
\frac{5x}{x^2-x-2} = \frac{10}{3(x-2)} + \frac{5}{3(x+1)}\text{.}
\end{equation*}

While there is much to be enthusiastic about regarding CAS programs such as *WolframAlpha*, there are several things we should be cautious about: (1) a CAS only responds to exactly what is input; (2) a CAS can answer using powerful functions from very advanced mathematics; and (3) there are problems that even a CAS cannot do without additional human insight.

Although (1) likely goes without saying, we have to be careful with our input: if we enter syntax that defines the wrong function, the CAS will work with precisely the function we define. For example, if we are interested in evaluating the integral

\begin{equation*}
\int \frac{1}{16-5x^2} \, dx\text{,}
\end{equation*}

and we mistakenly enter

`integrate 1/16 - 5x^2 dx`

a CAS will (correctly) reply with

\begin{equation*}
\frac{1}{16}x - \frac{5}{3} x^3\text{.}
\end{equation*}

But if we are sufficiently well-versed in antidifferentiation, we will recognize that this function cannot be the one that we seek: integrating a rational function such as \(\frac{1}{16-5x^2}\text{,}\) we expect the logarithm function to be present in the result.

Regarding (2), even for a relatively simple integral such as \(\int \frac{1}{16-5x^2} \, dx\text{,}\) some CASs will invoke advanced functions rather than simple ones. For instance, if we use *Maple* to execute the command

`int(1/(16-5*x^2), x);`

the program responds with

\begin{equation*}
\int \frac{1}{16-5x^2} \, dx = \frac{\sqrt{5}}{20} \arctanh \left(\frac{\sqrt{5}}{4}x\right)\text{.}
\end{equation*}

While this is correct (save for the missing arbitrary constant, which *Maple* never reports), the inverse hyperbolic tangent function is not a common nor familiar one; a simpler way to express this function can be found by using the partial fractions method, and happens to be the result reported by *WolframAlpha*:

\begin{equation*}
\int \frac{1}{16-5x^2} \, dx = \frac{1}{8\sqrt{5}} \left(\log(4\sqrt{5}+5x) - \log(4\sqrt{5}-5x)\right) + \text{constant}\text{.}
\end{equation*}

Using sophisticated functions from more advanced mathematics is sometimes the way a CAS says to the user “I don’t know how to do this problem.” For example, if we want to evaluate

\begin{equation*}
\int e^{-x^2} \, dx\text{,}
\end{equation*}

and we ask *WolframAlpha* to do so, the input

`integrate exp(-x^2) dx`

results in the output

\begin{equation*}
\int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\erf (x) + \text{constant}\text{.}
\end{equation*}

The function “erf\((x)\)” is the *error function*, which is actually defined by an integral:

\begin{equation*}
\erf (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt\text{.}
\end{equation*}

So, in producing an output involving an integral, the CAS has basically reported back to us the very question we asked.

Finally, as remarked at (3) above, there are times that a CAS will actually fail without some additional human insight. If we consider the integral

\begin{equation*}
\int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx
\end{equation*}

and ask *WolframAlpha* to evaluate

`int (1+x) * exp(x) * sqrt(1+x^2 * exp(2x)) dx`

,

the program thinks for a moment and then reports

(*no result found in terms of standard mathematical functions*)

But in fact this integral is not that difficult to evaluate. If we let \(u = xe^{x}\text{,}\) then \(du = (1+x)e^x \, dx\text{,}\) which means that the preceding integral has the form

\begin{equation*}
\int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx = \int \sqrt{1+u^2} \, du\text{,}
\end{equation*}

which is a straightforward one for any CAS to evaluate.

So, we should proceed with some caution: while any CAS is capable of evaluating a wide range of integrals (both definite and indefinite), there are times when the result can mislead us. We must think carefully about the meaning of the output, whether it is consistent with what we expect, and whether or not it makes sense to proceed.