7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
7.6
Integration Using Tables
and Computer Algebra Systems
In this section, we will learn:
How to use tables and computer algebra systems in
integrating functions that have elementary antiderivatives.
TABLES & COMPUTER ALGEBRA SYSTEMS
However, you should bear in mind that
even the most powerful computer algebra
systems (CAS) can’t find explicit formulas
for:
 The antiderivatives of functions like
2
x
e
 The other functions 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.
 We don’t have access to a CAS.
TABLES OF INTEGRALS
A relatively brief table of 120 integrals,
categorized by form, is provided on
the Reference Pages.
TABLES OF INTEGRALS
More extensive tables are available in:
 CRC Standard Mathematical Tables and Formulae,
31st ed. by Daniel Zwillinger (Boca Raton, FL: CRC
Press, 2002), which has 709 entries
 Gradshteyn and Ryzhik’s Table of Integrals, Series,
and Products, 6e (San Diego: Academic Press, 2000),
which contains hundreds of pages of integrals
TABLES OF INTEGRALS
Remember, integrals do not often occur
in exactly the form listed in a table.
 Usually, we need to use substitution or algebraic
manipulation to transform a given integral into
one of the forms in the table.
TABLES OF INTEGRALS
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.
TABLES OF INTEGRALS
Example 1
Using the method of cylindrical shells,
we see that the volume is:
1
V   2 x arctan x dx
0
Example 1
TABLES OF INTEGRALS
In the section of the Table of Integrals
titled Inverse Trigonometric Forms,
we locate Formula 92:
u  1 1
u
u
tan
u
du

tan
u


C

u
2
1
2
Example 1
TABLES OF INTEGRALS
1
So, the volume is: V  2 x tan 1 x dx

0
1
 x  1 1
x
 2 
tan x  
2 0
 2
2
   ( x  1) tan x  x 
1
2
  (2 tan 1  1)
  [2( / 4)  1]
1
  
1
2
2
1
0
TABLES OF INTEGRALS
Example 2
Use the Table of Integrals to find

x
2
5  4x
 If we look at the section of the table titled
‘Forms involving a 2  u 2 ,’ we see that
the closest entry is number 34:

2
u 2
a
2
1  u 
du  
a  u  sin    C
2
2
a
a2  u 2
u2
2
dx
Example 2
TABLES OF INTEGRALS
 That is not exactly what we have.
 Nevertheless, we will be able to use it if we first
make the substitution u = 2x:

x2
5  4x
2
dx  
(u / 2) 2 du
2 2
5u
1
u2
 
du
8 5  u2
TABLES OF INTEGRALS
Example 2
Then, we use Formula 34 with a2 = 5
(so a  5 ):

x2
1
u2
dx  
du
8 5  u2
5  4 x2
1 u
5 1 u 
2
 
5  u  sin
C

8 2
2
5
x
5
2
1  2 x 

5  4 x  sin 
C

8
16
 5
TABLES OF INTEGRALS
Example 3
Use the Table of Integrals to find
x
3
sin x dx
 If we look in the section Trigonometric Forms, we see
that none of the entries explicitly includes a u3 factor.
 However, we can use the reduction formula in entry 84
3
3
2
with n = 3:
x
sin x dx   x cos x  3 x cos x dx
TABLES OF INTEGRALS
Example 3
Now, we need to evaluate  x cos x dx
2
 We can use the reduction formula in entry 85
n
n
n 1
u
cos
u
du

u
sin
u

n
u

 sin u du
with n = 2.
 Then, we follow by entry 82:
x
2
cos x dx  x sin x  2 x sin x dx
2
 x 2 sin x  2(sin x  x cos x)  K
TABLES OF INTEGRALS
Example 3
Combining these calculations, we get
x
3
sin x dx   x cos x  3x sin x
3
2
 6 x cos x  6sin x  C
where C = 3K
Example 4
TABLES OF INTEGRALS
Use the Table of Integrals to find
x
x  2 x  4 dx
2
 The table gives forms involving a2  x2 , a2  x2 ,
and x2  a2 , but not ax 2  bx  c .
 So, we first complete the square:
x 2  2 x  4  ( x  1)2  3
Example 4
TABLES OF INTEGRALS
If we make the substitution u = x + 1
(so x = u – 1), the integrand will involve
the pattern a2  u 2 :
x
x  2 x  4 dx   (u  1) u  3 du
2
2
  u u  3 du   u  3 du
2
2
Example 4
TABLES OF INTEGRALS
The first integral is evaluated using
the substitution t = u2 + 3:
u
u

3
du


2
1
2

t dt
  t
1
2
2 3/ 2
3
 (u  3)
1
3
2
3/ 2
TABLES OF INTEGRALS
Example 4
For the second integral, we use the formula

2
u
a
a 2  u 2 du 
a 2  u 2  ln(u  a 2  u 2 )  C
2
2
with a  3 :

u 2
2
3
u  3 du 
u  3  2 ln(u  u  3)
2
2
Example 4
TABLES OF INTEGRALS
Thus,
x
x  2 x  4 dx
2
 ( x  2 x  4)
1
3
2
32
x 1 2

x  2x  4
2
 ln( x  1  x  2 x  4)  C
3
2
2
COMPUTER ALGEBRA SYSTEMS
We have seen that the use of tables
involves matching the form of the given
integrand with the forms of the integrands
in the tables.
CAS
Computers are particularly good at matching
patterns.
Also, just as we used substitutions in
conjunction with tables, a CAS can perform
substitutions that transform a given integral
into one that occurs in its stored formulas.
 So, it isn’t surprising that CAS excel at integration.
CAS
That doesn’t mean that integration by
hand is an obsolete skill.
 We will see that, sometimes, a hand computation
produces an indefinite integral in a form that is
more convenient than a machine answer.
CAS VS. MANUAL COMPUTATION
To begin, let’s see what happens when
we ask a machine to integrate the relatively
simple function
y = 1/(3x – 2)
CAS VS. MANUAL COMPUTATION
Using the substitution u = 3x – 2, an easy
calculation by hand gives:
1
1
dx

3 ln 3 x  2  C
 3x  2
However, Derive, Mathematica, and Maple
return:
1
3
ln(3x  2)
CAS VS. MANUAL COMPUTATION
The first thing to notice is that CAS omit
the constant of integration.
 That is, they produce a particular antiderivative,
not the most general one.
 Thus, when making use of a machine integration,
we might have to add a constant.
CAS VS. MANUAL COMPUTATION
Second, the absolute value signs are
omitted in the machine answer.
 That is fine if our problem is concerned only
with values of x greater than 2 .
3
 However, if we are interested in other values of x,
then we need to insert the absolute value symbol.
CAS
In the next example, we reconsider the
integral of Example 4.
This time, though, we ask a machine for
the answer.
Example 5
CAS
Use a CAS to find  x x 2  2 x  4 dx
 Maple responds with:
1
3
( x 2  2 x  4)3 2  14 (2 x  2) x 2  2 x  4
3
3
 arcsinh
(1  x)
2
3
CAS
Example 5
That looks different from the answer in
Example 4.
However, it is equivalent because the third
term can be rewritten using the identity
arcsinh x  ln( x  x 2  1)
Example 5
CAS
Thus,
 3
3
arc sinh
(1  x)  ln  (1  x) 
3
 3
1
3

(1  x)  1 

2
1 
 ln
1  x  (1  x) 2  3 

3
1
2
 ln
 ln x  1  x  2 x  4
3



 The resulting extra term  32 ln 1/ 3 can be absorbed
into the constant of integration.

CAS
Example 5
Mathematica gives:
 5 x x2  2
3
 1 x 
    x  2 x  4  arcsinh 

2
 3 
6 6 3 
 It combined the first two terms of Example 4
(and the Maple result) into a single term by factoring.
Example 5
CAS
Derive gives:
1
6
x  2 x  4 (2 x  x  5)
2
2
 ln
3
2


x  2x  4  x  1
2
 The first term is like the first term in the Mathematica
answer.
 The second is identical to the last term in Example 4.
Example 6
CAS

Use a CAS to evaluate x ( x  5) dx
2
8
 Maple and Mathematica give the same answer:
1
18
12
10
x18  52 x16  50 x14  1750
x

4375
x
3
 21875 x 
8
218750
3
x  156250 x 
6
4
390625
2
x
2
CAS
Example 6
It’s clear that both systems
must have expanded (x2 + 5)8
by the Binomial Theorem and then
integrated each term.
Example 6
CAS
If we integrate by hand instead, using
the substitution u = x2 + 5, we get:
 x( x
2
 5) dx 
8
1
18
( x  5)  C
2
9
 For most purposes, this is a more convenient
form of the answer.
E. g. 7—Equation 1
CAS
Use a CAS to find  sin 5 x cos 2 x dx
 In Example 2 in Section 7.2, we found:
 sin
5
2
x cos x dx
 13 cos3 x  52 cos5 x  71 cos 7 x  C
Example 7
CAS
Derive and Maple report:
 sin x cos x  sin x cos x 
1
7
4
3
4
35
2
3
8
105
3
cos x
Mathematica produces:
3
1
1
 645 cos x   192
cos3x  320
cos5 x   448
cos 7 x
CAS
Example 7
We suspect there are trigonometric
identities that show these three answers
are equivalent.
 Indeed, if we ask Derive, Maple, and Mathematica
to simplify their expressions using trigonometric
identities, they ultimately produce the same form
of the answer as in Equation 1.