514
CHAPTER 6
.
.
Integration Techniques
6-6
46.
±
sec
xdx
and
±
sec
2
47.
Find
±
2
0
f
(
x
)
dx
, where
f
(
x
)
=
²
x
/
(
x
2
+
1)
if
x
≤
1
x
2
/
(
x
2
+
1)
if
x
>
1
48.
Find
±
2
−
2
f
(
x
)
, where
f
(
x
)
=
²
xe
x
2
if
x
<
0
x
2
e
x
3
if
x
≥
0
49.
Rework
example
1.5
by
rewriting
the
integral
as
±
4
x
+
4
2
x
2
+
4
x
+
10
−
±
3
2
x
2
+
4
x
+
10
and complet-
ing the square in the second integral.
EXPLORATORY EXERCISES
1.
Find
±
1
1
+
x
2
,
±
x
1
+
x
2
,
±
x
2
1
+
x
2
and
±
x
3
1
+
x
2
.
Generalize to give the form of
±
x
n
1
+
x
2
for any positive
integer
n
,as completely as you can.
2.
Find
±
x
1
+
x
4
,
±
x
3
1
+
x
4
and
±
x
5
1
+
x
4
. Generalize
to give the form of
±
x
n
1
+
x
4
for any odd positive integer
n
.
3.
Use a CAS to ﬁnd
³
−
x
2
,
³
x
3
e
−
x
2
and
³
x
5
e
−
x
2
.
Verify that each antiderivative is correct. Generalize to give
the form of
³
x
n
e
−
x
2
for any odd positive integer
n
.
4.
In
many
situations,
the
integral
as
we’ve
deﬁned
it
must be extended to the
Riemann-Stieltjes integral
con-
sidered in this exercise. For functions
f
and
g
, let
P
be a regular partition of [
a
,
b
] and deﬁne the sums
R
(
f
,
g
,
P
)
=
n
∑
i
=
1
f
(
c
i
)[
g
(
x
i
)
−
g
(
x
i
−
1
)].
The
integral
³
b
a
f
(
x
)
dg
(
x
) equals the limit of the sums
R
(
f
,
g
,
P
)
as
n
→∞
,i
f
the
limit
exists
and
equals
the
same
number for all evaluation points
c
i
. (a) Show that if
g
±
exists,
then
³
b
a
f
(
x
)
(
x
)
=
³
b
a
f
(
x
)
g
±
(
x
)
.
(b)
If
g
(
x
)
=
´
1
a
≤
x
≤
c
2
c
<
x
≤
b
for some constant
c
with
a
<
c
<
b
,
evaluate
³
b
a
f
(
x
)
(
x
). (c) Find a function
g
(
x
) such that
³
1
0
1
x
(
x
)exists.
6.2
INTEGRATION BY PARTS
At this point, you will have recognized that there are many integrals that cannot be evaluated
using our basic formulas or integration by substitution. For instance,
±
x
sin
cannot be evaluated with what you presently know. We improve this situation in the current
section by introducing a powerful tool called
integration by parts
.
HISTORICAL NOTES
Brook Taylor (1685–1731)
An English mathematician who is
credited with devising integration
by parts. Taylor made important
contributions to probability, the
theory of magnetism and the use
of vanishing lines in linear perspec-
tive. However, he is best known
for Taylor’s Theorem (see section
8.7), in which he generalized
results of Newton, Halley, the
Bernoullis and others. Personal
tragedy (both his wives died
during childbirth) and poor health
limited the mathematical output
of this brilliant mathematician.
We have observed that every differentiation rule gives rise to a corresponding integration
rule. So, for the product rule:
d
[
f
(
x
)
g
(
x
)]
=
f
±
(
x
)
g
(
x
)
+
f
(
x
)
g
±
(
x
)
,
integrating both sides of this equation gives us
±
d
[
f
(
x
)
g
(
x
)]
=
±
f
±
(
x
)
g
(
x
)
+
±
f
(
x
)
g
±
(
x
)
.
Ignoring the constant of integration, the integral on the left-hand side is simply
f
(
x
)
g
(
x
).
Solving for the second integral on the right-hand side then yields
±
f
(
x
)
g
±
(
x
)
=
f
(
x
)
g
(
x
)
−
±
f
±
(
x
)
g
(
x
)
.