CHAPTER 7
Techniques of Integration
7.1. Substitution
Integration, unlike differentiation, is more of an artform than a collection of algorithms. Many problems
in applied mathematics involve the integration of functions given by complicated formulae, and practi
tioners consult a
Table of Integrals
in order to complete the integration. There are certain methods of
integration which are essential to be able to use the Tables effectively. These are: substitution, integration
by parts and partial fractions. In this chapter we will survey these methods as well as some of the ideas
which lead to the tables. After the examination on this material, students will be free to use the Tables to
integrate.
The idea of substitution was introduced in section 4.1 (recall Proposition 4.4). To integrate a differen
tial
f x dx
which is not in the table, we first seek a function
u
u x
so that the given differential can be
rewritten as a differential
g u du
which does appear in the table. Then, if
g u du
G u
C
, we know
that
f x dx
G u x
C
. Finding and employing the function
u
often requires some experience and
ingenuity as the following examples show.
Example 7.1
x
2
x
1
dx
?
Let
u
2
x
1, so that
du
2
dx
and
x
u
1
2. Then
x
2
x
1
dx
u
1
2
u
1 2
du
2
1
4
u
3 2
u
1 2
du
1
4
2
5
u
5 2
2
3
u
3 2
C
(7.1)
1
30
u
3 2
3
u
5
C
1
30
2
x
1
3 2
6
x
2
C
1
15
2
x
1
3 2
3
x
1
C
(7.2)
where at the end we have replaced
u
by 2
x
1.
Example 7.2
tan
xdx
?
107
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Chapter 7
Techniques of Integration
108
Since this isn’t on our tables, we revert to the definition of the tangent: tan
x
sin
x
cos
x
. Then,
letting
u
cos
x du
sin
xdx
we obtain
(7.3)
tan
xdx
sin
x
cos
x
dx
du
u
ln
u
C
lncos
x
C
lnsec
x
C
Example 7.3
sec
xdx
?.
This is tricky, and there are several ways to find the integral.
However, if we are guided by the
principle of rewriting in terms of sines and cosines, we are led to the following:
(7.4)
sec
x
1
cos
x
cos
x
cos
2
x
cos
x
1
sin
2
x
Now we can try the substitution
u
sin
x du
cos
xdx
. Then
(7.5)
sec
xdx
du
1
u
2
This looks like a dead end, but a little algebra pulls us through. The identity
(7.6)
1
1
u
2
1
2
1
1
u
1
1
u
leads to
(7.7)
du
1
u
2
dx
1
2
1
1
u
1
1
u
du
1
2
ln 1
u
ln 1
u
C
Using
u
sin
x
, we finally end up with
(7.8)
sec
xdx
1
2
ln 1
sin
x
ln 1
sin
x
C
1
2
ln
1
sin
x
1
sin
x
C
Example 7.4
As a circle rolls along a horizontal line, a point on the circle traverses a curve called the
cycloid
. A
loop
of the cycloid is the trajectory of a point as the circle goes through one full rotation. Let
us find the length of one loop of the cycloid traversed by a circle of radius 1.
Let the variable
t
represent the angle of rotation of the circle, in radians, and start (at
t
0) with the
point of intersection
P
of the circle and the line on which it is rolling. After the circle has rotated through
t
radians, the position of the point is as given as in figure 7.1. The point of contact of the circle with the
line is now
t
units to the right of the original point of contact (assuming no slippage), so
(7.9)
x t
t
sin
t
y t
1
cos
t
To find arc length, we use
ds
2
dx
2
dy
2
, where
dx
1
cos
t dt
dy
sin
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 Spring '08
 BLOCK
 Calculus, Applied Mathematics

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