Integrals Yielding Logarithmic and Exponential
Functions
For today
1
Integrals of
f
(
x
) =
1
x
and of the other Circular Functions
2
Integrals of Exponential Functions
Recall:
Z
u
n
du
=
u
n
+
1
n
+
1
+
C
whenever
n
6
=
−
1
Theorem
Z
1
u
du
=
ln
|
u
|
+
C
This is true since
Du
(ln
|
u
|
) =
1
Theorem
Z
1
u
du
=
ln
|
u
|
+
C
Examples.
Evaluate the following integrals.
1.
Z
6
3
x
+
2
dx
Let
u
=
3
x
+
2
du
=
3
dx
=
Z
2
u
du
=
2 ln
|
u
|
+
C
Theorem
Z
1
u
du
=
ln
|
u
|
+
C
2.
Z
4
3
x
4
−
x
2
dx
Let
u
=
4
−
x
2
if
x
=
3
u
=
−
5
du
=
−
2
xdx
if
x
=
4
u
=
−
12
=
−
1
2
Z
−
12
−
5
1
u
du
=
−
1
2
ln
|
u
|
u
=
−
12
u
=
−
5
=
−
1
2
ln
| −
12
|
−
−
1
2
ln
| −
5
|
=
ln
r
Theorem
Z
1
u
du
=
ln
|
u
|
+
C
3.
Z
1
x
log
2
x
dx
Let
u
=
log
2
x
du
=
1
x
ln 2
dx
=
ln 2
Z
1
u
du
= (ln 2)
ln
|
u
|
+
C
4.
Z
2
x
3
+
9
x
2
−
3
x
+
2
2
x
−
1
dx
=
Z
x
2
+
5
x
+
1
+
3
2
x
−
1
dx
using long division
=
Z
(
x
2
+
5
x
+
1)
dx
+
Z
3
2
x
−
1
dx
Let
u
=
2
x
−
1
du
=
2
dx
=
Z
(
x
2
+
5
x
+
1)
dx
+
3
2
Z
1
u
du
=
x
3
3
+
5
x
2
2
+
x
+
3
2
ln
|
u
|
+
C
=
x
3
3
+
5
x
2
2
+
x
+
3
5.
Z
x
−
1
x
2
+
2
x
+
1
dx
=
Z
x
−
1
(
x
+
1
)
2
dx
Let
u
=
x
+
1
du
=
dx
x
−
1
=
u
−
2
=
Z
u
−
2
u
2
du
=
Z
1
u
−
2
u
2
du
=
Z
1
u
−
2
Z
1
u
2
du
=
ln
|
u
|
+
2
u
+
C
Integrals of the other Circular Functions
Recall: We know what
Z
sin
u du
and
Z
cos
u du
are.
1
Z
tan
u du
=?
2
Z
cot
u du
=?
3
Z
sec
u du
=?
4
Z
Z
tan
u du
Z
tan
u du
=
Z
sin
u
cos
u
du
Let
v
=
cos
u
dv
=
−
sin
u du
=
−
Z
1
v
dv
=
−
ln
|
v
|
+
C
=
−
ln
|
cos
u
|
+
C
Integrals of the other Circular Functions
Theorem
1
Z
tan
u du
=
ln
|
sec
u
|
+
C
=
−
ln
|
cos
u
|
+
C
2
Z
cot
u du
=
ln
|
sin
u
|
+
C
3
Z
sec
u du
=
ln
|
sec
u
+
tan
u
|
+
C
4
Z
Z
sec
u du
=
ln
|
sec
u
+
tan
u
|
+
C
Proof.
Z
sec
u du
=
Z
sec
u
(
sec
u
+
tan
u
)
sec
u
+
tan
u
du
=
Z
sec
2
u
+
sec
u
tan
u
sec
u
+
tan
u
du
Let
v
=
sec
u
+
tan
u
dv
= (sec
u
tan
u
+
sec
2
u
)
du
=
Z
dv
v
=
ln
|
v
|
+
C
Theorem
1
Z
tan
u du
=
ln
|
sec
u
|
+
C
2
Z
cot
u du
=
ln
|
sin
u
|
+
C
3
Z
sec
u du
=
ln
|
sec
u
+
tan
u
|
+
C
4
Z
csc
u du
=
ln
|
csc
u
−
cot
u
|
+
C
Examples.
Evaluate the following integrals.
1.
Z
x
3
csc
x
4
dx
Let
u
=
x
4
du
=
4
x
3
dx
=
1
4
Z
csc
u du
=
1
4
ln
|
csc
u
−
cot
u
|
+
C
=
1
4
ln
csc
Theorem
1
Z
tan
u du
=
ln
|
sec
u
|
+
C
2
Z
cot
u du
=
ln
|
sin
u
|
+
C
3
Z
sec
u du
=
ln
|
sec
u
+
tan
u
|
+
C
4
Z
csc
u du
=
ln
|
csc
u
−
cot
u
|
+
C
2.
Z
23ππ 2
sin(
x
2
) +
4
cos
(
x
2
)
dx
Let
u
=
x
2
if
x
=
π
2
u
=
π
4
du
=
1
2
dx
if
x
=
2
π
3
u
=
π
3
=
2
Z
π3π 4
sin
u
+
4
cos
u
du
=
2
Z
π3π 4
(tan
u
+
4 sec
u
)
du
=
2
(ln
|
sec
u
|
+
4 ln
|
sec
u
+
tan
u
|
)
u
=
π 3u
=
π 4Integrals of Exponential Functions
Theorem
1
Z
e
u
du
=
e
u
+
C
2
Z
Theorem
1
Z
e
u
du
=
e
u
+
C
2
Z
a
u
du
=
a
u
ln
a
+
C
Examples.
Evaluate the following integrals.
1.
Z
7
x
dx
Z
Theorem
1
Z
e
u
du
=
e
u
+
C
2
Z
a
u
du
=
a
u
ln
a
+
C
2.
Z
e
1
−
4
x
dx
Let
u
=
1
−
4
x
du
=
−
4
dx
=
−
1
4
Z
e
u
du
=
−
1
4
e
u
+
C
=
−
e
Theorem
1
Z
e
u
du
=
e
u
+
C
2
Z
a
u
du
=
a
u
ln
a
+
C
3.
Z
2
2
x
·
5
x
dx
Z
2
2
x
·
5
x
dx
=
Z
4
x
·
5
x
dx
=
Z
Theorem
1
Z
e
u
du
=
e
u
+
C
2
Z
a
u
du
=
a
u
ln
a
+
C
4.
Z
x
2
·
2
5
−
x
3dx
Let
u
=
5
−
x
3
du
=
−
3
x
2
dx
=
−
1
3
Z
2
u
du
=
−
1
3
·
2
u
ln 2
+
C
=
−
2
5.
Z
e
1
1
x
·
2
ln
x
dx
Let
u
=
ln
x
if
x
=
1
u
=
0
du
=
1
x
dx
if
x
=
e
u
=
1
=
Z
1
0
1
2
u
du
=
Z
1
0
1
2
u
du
=
1
2
u
ln
1
2
u
=
1
u
=
0
=
1
2
ln
1
2
−
1
ln
1
2
=
−
1
6.
Z
e
x
e
x
+
e
−
x
dx
Z
e
x
e
x
+
e
−
x
·
e
x
e
x
dx
=
Z
e
2
x
e
2
x
+
1
dx
Let
u
=
e
2
x
+
1
du
=
2
e
2
x
dx
=
1
2
Z
1
u
du
=
1
2
ln
|
u
|
+
C
=
1
2
ln
|
e
Exercises
1
Z
2
−
x
x
2
−
4
x
dx
2
Z
log(
x
3
)
x
dx
3
Z
sin(2
x
)
cos(2
x
)
1
+
cos(4
x
)
dx
4
Z
5
cot
x
−
sin
x
Recall: Let
a
>
0
,
n
∈
N
and
m
∈
Z
.
a
n
=
a
·
a
·
a
· · ·
a
|
{z
}
n
factors
a
0
=
1
a
−
n
=
1
a
n
a
m/
n= (
√
na
)
m
A Different Approach
Consider
f
(
t
) =
1
t
,
t
>
0
.
Then
f
is continuous on
(0,
+
∞
)
.
Consider
F
(
x
) =
Z
x
1
1
t
dt
,
x
>
0
.
f
(
t
) =
1F
(
x
) =
Z
x
1
1
t
dt
f
(
t
) =
1t
1
x
R
x
R
If
x
>
1
,
F
(
x
) =
Area of R
If
0
<
x
<
1
,
F
(
x
) =
−
Area of R
F
0
(
x
) =
1
x
(FTOC I)
Definition
The
natural logarithmic function
, denoted by
ln
, is defined by
ln
x
=
Z
x
1
1
t
dt
ln
x
=
Z
x
1
1
t
dt
Properties
Let
a
,
b
>
0
and
r
∈
Q
.
1Dx
(ln
x
) =
1
x
2
ln 1
=
0
3
ln
ab
=
ln
a
+
ln
b
4
ln
a
b
=
ln
a
−
ln
b
ln
ab
=
ln
a
+
ln
b
Proof.
ln
ab
=
Z
ab
1
1
t
dt
=
Z
a
1
1
t
dt
+
Z
ab
a
1
t
dt
Let
u
=
t
a
du
=
1
a
dt
if
t
=
a
u
=
1
if
t
=
ab
u
=
b
=
Z
a
1
1
t
dt
+
Z
b
1
1
u
du
Graph of
F
(
x
) =
ln
x
F
(
x
) =
ln
x
x
>
0
F
0
(
x
) =
1
x
>
0
for all
x
>
0
F
00
(
x
) =
−
1
x
2
<
0
for all
x
>
0
lim
x
→
0
+
F
(
x
) =
−
∞
lim
x
→
+
∞
F
(
x
) = +
∞
F
(1) =
0
Defining
e
There exists a real number
a
such that
ln
a
=
1
.
f
(
t
) =
1t
1
a
1
F
(
x
) =
ln
x
a
1
Definition
The number
e
is defined as the unique real number such that
Z
e
Definition
The
natural exponential function
, denoted by
exp
, is the inverse of the natural
logarithmic function.
Remark
1
y
=
ln
x
if and only if
x
=
exp(
y
)
2dom exp
=
ran ln
=
R
3
ran exp
=
dom ln
= (0,
+
∞
)
4
ln
e
=
1
⇒
exp(1) =
e
5ln(exp
x
) =
x
for all
x
∈
R
6exp(ln
x
) =
x
for all
x
∈
(0,
+
∞
)
Definition
1
If
a
>
0
and
x
∈
R
, then the
x
th power of
a
is defined by
a
x
=
exp(
x
ln
a
)
2
If
x
>
0
, we define
0
x
=
0
.
Thus,
e
x
=
exp(
x
ln
e
) =
exp(
x
)
.
Remark
Definition
Let
a
>
0
and
a
6
=
1
.
1
The
exponential function with base
a
is
f
(
x
) =
a
x
=
e
x
ln
a
.
2The
logarithmic function with base
a
, denoted by
log
a
, is the inverse
function of
f
(
x
) =
a
x
.
From these definitions, one can
1
prove laws of exponents and logarithms
Exercises
Perform the integration.
1Z
2
−
x
x
2
−
4
x
dx
2
Z
log
4
(
x
2
)
x
dx
3
Z
1
x
2
e
1xdx
4
Z
sin(2
x
)
cos(2
x
)
1
+
cos(4
x
)
dx
5
Z
3
−
√
x
2
√
x
√
x
+
3
√
x
sin(2
x
)
!