James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 7.6 Integration Using Tables and Computer Algebra Systems 509
the most powerful computer algebra systems can’t find explicit formulas for the antiderivatives
of functions like e x 2 or the other functions described at the end of Section 7.5.
Tables of Integrals
Tables of indefinite integrals are very useful when we are confronted by an integral
that is difficult to evaluate by hand and we don’t have access to a computer algebra
system. A relatively brief table of 120 integrals, categorized by form, is provided on
the Reference Pages at the back of the book. More extensive tables are available in the
CRC Standard Mathe matical Tables and Formulae, 31st ed. by Daniel Zwillinger (Boca
Raton, FL, 2002) (709 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series,
and Products, 7e (San Diego, 2007), which contains hundreds of pages of integrals. It
should be remembered, however, that integrals do not often occur in exactly the form
listed in a table. Usually we need to use the Substitution Rule or algebraic manipulation
to transform a given integral into one of the forms in the table.
Example 1 The region bounded by the curves y − arctan x, y − 0, and x − 1 is
rotated about the y-axis. Find the volume of the resulting solid.
SOLUtion Using the method of cylindrical shells, we see that the volume is
V − y 1
2x arctan x dx
0
The Table of Integrals appears on
Reference Pages 6–10 at the back
of the book.
In the section of the Table of Integrals titled Inverse Trigonometric Forms we locate
Formula 92:
y u tan 21 u du − u 2 1 1
tan 21 u 2 u 2
2 1 C
So the volume is
V − 2 y 1
0
x tan 21 x dx − 2F x 2 1 1
2
1
tan 21 x 2 x 2G0
− fsx 2 1 1d tan 21 x 2 xg 0
1
− s2 tan
21
1 2 1d
− f2sy4d 2 1g − 1 2 2 2 n
Example 2 Use the Table of Integrals to find y
x 2
s5 2 4x 2 dx.
SOLUtion If we look at the section of the table titled Forms Involving sa 2 2 u 2 ,
we see that the closest entry is number 34:
y
u 2
sa 2 2 u 2 du − 2 u 2 sa 2 2 u 2 1 a 2
2 sin21S u aD 1 C
This is not exactly what we have, but we will be able to use it if we first make the substitution
u − 2x:
y
x 2
s5 2 4x 2 dx − y
suy2d 2
s5 2 u 2
du
2 − 1 8 y
u 2
s5 2 u 2 du
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