5.5. OTHER OPTIONS FOR FINDING ALGEBRAIC ANTIDERIVATIVES
313
Using an Integral Table
Calculus has a long history, with key ideas going back as far as Greek mathematicians in
400-300 BC. Its main foundations were first investigated and understood independently by
Isaac Newton and Gottfried Wilhelm Leibniz in the late 1600s, making the modern ideas
of calculus well over 300 years old. It is instructive to realize that until the late 1980s, the
personal computer essentially did not exist, so calculus (and other mathematics) had to be
done by hand for roughly 300 years. During the last 30 years, however, computers have
revolutionized many aspects of the world we live in, including mathematics. In this section
we take a short historical tour to precede the following discussion of the role computer
algebra systems can play in evaluating indefinite integrals. In particular, we consider a
class of integrals involving certain radical expressions that, until the advent of computer
algebra systems, were often evaluated using an integral table.
As seen in the short table of integrals found in Appendix A, there are also many forms
of integrals that involve
√
a 2 ± w 2 and
√
w 2 − a 2 . These integral rules can be developed
using a technique known as trigonometric substitution that we choose to omit; instead, we
will simply accept the results presented in the table. To see how these rules are needed
and used, consider the differences among
1
√
1 − x 2
dx,
x
√
1 − x 2
dx, and
√
1 − x 2 dx.
The first integral is a familiar basic one, and results in arcsin(x) + C. The second integral
can be evaluated using a standard u-substitution with u = 1 − x 2 . The third, however, is
not familiar and does not lend itself to u-substitution.
In Appendix A, we find the rule
(8)
√
a 2 − u 2 du =
u
2
√
a 2 − u 2 +
a 2
2
arcsin
u
a
+ C.
Using the substitutions a = 1 and u = x (so that du = dx), it follows that
√
1 − x 2 dx =
x
2
√
1 − x 2 −
1
2
arcsin x + C.
One important point to note is that whenever we are applying a rule in the table, we
are doing a u-substitution. This is especially key when the situation is more complicated
than allowing u = x as in the last example. For instance, say we wish to evaluate the
integral
√
9 + 64x 2 dx.
Once again, we want to use Rule (3) from the table, but now do so with a = 3 and u = 8x;
we also choose the “+” option in the rule. With this substitution, it follows that du = 8dx,
313
Using an Integral Table
Calculus has a long history, with key ideas going back as far as Greek mathematicians in
400-300 BC. Its main foundations were first investigated and understood independently by
Isaac Newton and Gottfried Wilhelm Leibniz in the late 1600s, making the modern ideas
of calculus well over 300 years old. It is instructive to realize that until the late 1980s, the
personal computer essentially did not exist, so calculus (and other mathematics) had to be
done by hand for roughly 300 years. During the last 30 years, however, computers have
revolutionized many aspects of the world we live in, including mathematics. In this section
we take a short historical tour to precede the following discussion of the role computer
algebra systems can play in evaluating indefinite integrals. In particular, we consider a
class of integrals involving certain radical expressions that, until the advent of computer
algebra systems, were often evaluated using an integral table.
As seen in the short table of integrals found in Appendix A, there are also many forms
of integrals that involve
√
a 2 ± w 2 and
√
w 2 − a 2 . These integral rules can be developed
using a technique known as trigonometric substitution that we choose to omit; instead, we
will simply accept the results presented in the table. To see how these rules are needed
and used, consider the differences among
1
√
1 − x 2
dx,
x
√
1 − x 2
dx, and
√
1 − x 2 dx.
The first integral is a familiar basic one, and results in arcsin(x) + C. The second integral
can be evaluated using a standard u-substitution with u = 1 − x 2 . The third, however, is
not familiar and does not lend itself to u-substitution.
In Appendix A, we find the rule
(8)
√
a 2 − u 2 du =
u
2
√
a 2 − u 2 +
a 2
2
arcsin
u
a
+ C.
Using the substitutions a = 1 and u = x (so that du = dx), it follows that
√
1 − x 2 dx =
x
2
√
1 − x 2 −
1
2
arcsin x + C.
One important point to note is that whenever we are applying a rule in the table, we
are doing a u-substitution. This is especially key when the situation is more complicated
than allowing u = x as in the last example. For instance, say we wish to evaluate the
integral
√
9 + 64x 2 dx.
Once again, we want to use Rule (3) from the table, but now do so with a = 3 and u = 8x;
we also choose the “+” option in the rule. With this substitution, it follows that du = 8dx,