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7.1: Integration by Parts
(Dated: October 12, 2011)
Let
f
(
x
) and
g
(
x
) be differentiable functions. Then the
product rule states that
d
dx
£
f
(
x
)
g
(
x
)
/
=
f
(
x
)
g
0
(
x
)
+
g
(
x
)
f
0
(
x
).
This shows that
f
(
x
)
g
0
(
x
)
+
g
(
x
)
f
0
(
x
) has
f
(
x
)
g
(
x
) as an an
tiderivative, from which it follows that
Z
f
(
x
)
g
0
(
x
)
dx
+
Z
g
(
x
)
f
0
(
x
)
dx
=
f
(
x
)
g
(
x
).
It is customary to write this formula as
Z
f
(
x
)
g
0
(
x
)
dx
=
f
(
x
)
g
(
x
)

Z
g
(
x
)
f
0
(
x
)
dx
.
It is called the
formula
for
integration
by
parts. In terms of
differentials, one has
Z
u dv
=
uv

Z
v du
,
where
u
=
f
(
x
) and
v
=
g
(
x
). For deﬁnite integrals,
Z
b
a
f
(
x
)
g
0
(
x
)
dx
=
£
f
(
x
)
g
(
x
)
/
b
a

Z
b
a
g
(
x
)
f
0
(
x
)
dx
.
Example 1 (Exercise 7.1.4 in the text)
Let u
=
x and v
0
=
e

x
. Then
u
0
=
1
and v
=
e

x
.
Thus, by integration by parts,
Z
xe

x
dx
=
xe

x

Z

e

x
dx
=
xe

x

e

x
+
C
.
Example 2 (Exercise 7.1.7 in the text)
Let u
=
x
2
and v
0
=
sin
π
x. Then
u
0
=
2
x and v
=
cos
π
x
π
.
Thus, by integration by parts,
Z
x
2
sin
π
x dx
=
x
2
cos
π
x
π

Z

2
x
cos
π
x
π
dx
=
x
2
cos
π
x
π
+
2
π
Z
x
cos
π
x dx
.
Let u
1
=
x and v
0
1
=
cos
π
x. Then
u
0
1
=
1
and v
1
=
sin
π
x
π
.
Then, by integration by parts also,
Z
x
cos
π
x dx
=
x
sin
π
x
π

Z
sin
π
x
π
dx
=
x
sin
π
x
π
+
cos
π
x
π
2
+
C
1
.
Consequently,
Z
x
2
sin
π
x dx
=
x
2
cos
π
x
π
+
2
π
±
x
sin
π
x
π
+
cos
π
x
π
2
+
C
1
¶
=
x
2
cos
π
x
π
+
2
x
sin
π
x
π
2
+
2cos
π
x
π
3
+
C
,
where C
=
2
C
1
π
.
Example 3 (Exercise 7.1.9 in the text)
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