514
CHAPTER 6
.
.
Integration Techniques
66
46.
sec
x dx
and
sec
2
x dx
47.
Find
2
0
f
(
x
)
dx
, where
f
(
x
)
=
x
/
(
x
2
+
1)
if
x
≤
1
x
2
/
(
x
2
+
1)
if
x
>
1
48.
Find
2
−
2
f
(
x
)
dx
, where
f
(
x
)
=
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
dx
−
3
2
x
2
+
4
x
+
10
dx
and complet
ing the square in the second integral.
EXPLORATORY EXERCISES
1.
Find
1
1
+
x
2
dx
,
x
1
+
x
2
dx
,
x
2
1
+
x
2
dx
and
x
3
1
+
x
2
dx
.
Generalize to give the form of
x
n
1
+
x
2
dx
for any positive
integer
n
, as completely as you can.
2.
Find
x
1
+
x
4
dx
,
x
3
1
+
x
4
dx
and
x
5
1
+
x
4
dx
. Generalize
to give the form of
x
n
1
+
x
4
dx
for any odd positive integer
n
.
3.
Use a CAS to find
xe
−
x
2
dx
,
x
3
e
−
x
2
dx
and
x
5
e
−
x
2
dx
.
Verify that each antiderivative is correct. Generalize to give
the form of
x
n
e
−
x
2
dx
for any odd positive integer
n
.
4.
In
many
situations,
the
integral
as
we’ve
defined
it
must be extended to the
RiemannStieltjes 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
b
a
f
(
x
)
dg
(
x
)
equals
the
limit
of
the
sums
R
(
f
,
g
,
P
)
as
n
→ ∞
,
if
the
limit
exists
and
equals
the
same
number
for
all
evaluation
points
c
i
.
(a)
Show
that
if
g
exists,
then
b
a
f
(
x
)
dg
(
x
)
=
b
a
f
(
x
)
g
(
x
)
dx
.
(b)
If
g
(
x
)
=
1
a
≤
x
≤
c
2
c
<
x
≤
b
for some constant
c
with
a
<
c
<
b
,
evaluate
b
a
f
(
x
)
dg
(
x
). (c) Find a function
g
(
x
) such that
1
0
1
x
dg
(
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
x dx
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.
Wehaveobservedthateverydifferentiationrulegivesrisetoacorrespondingintegration
rule. So, for the product rule:
d
dx
[
f
(
x
)
g
(
x
)]
=
f
(
x
)
g
(
x
)
+
f
(
x
)
g
(
x
)
,
integrating both sides of this equation gives us
d
dx
[
f
(
x
)
g
(
x
)]
dx
=
f
(
x
)
g
(
x
)
dx
+
f
(
x
)
g
(
x
)
dx
.
Ignoring the constant of integration, the integral on the lefthand side is simply
f
(
x
)
g
(
x
).
Solving for the second integral on the righthand side then yields
f
(
x
)
g
(
x
)
dx
=
f
(
x
)
g
(
x
)
−
f
(
x
)
g
(
x
)
dx
.
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 Spring '10
 atsumi
 Derivative, Sin

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