Tích phân hàm số toán cotangent
From testwiki
Revision as of 00:34, 5 July 2020 by
imported>JohnsonLee01
(Copying from
Category:Tích phân
to
Category:VI
using
Cat-a-lot
)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation
Jump to search
∫
cot
a
x
d
x
=
1
a
ln
|
sin
a
x
|
+
C
∫
cot
n
a
x
d
x
=
−
1
a
(
n
−
1
)
cot
n
−
1
a
x
−
∫
cot
n
−
2
a
x
d
x
(for
n
≠
1
)
∫
d
x
1
+
cot
a
x
=
∫
tan
a
x
d
x
tan
a
x
+
1
∫
d
x
1
−
cot
a
x
=
∫
tan
a
x
d
x
tan
a
x
−
1
∫
d
x
cos
a
x
±
sin
a
x
=
1
a
2
ln
|
tan
(
a
x
2
±
π
8
)
|
+
C
∫
d
x
(
cos
a
x
±
sin
a
x
)
2
=
1
2
a
tan
(
a
x
∓
π
4
)
+
C
∫
d
x
(
cos
x
+
sin
x
)
n
=
1
n
−
1
(
sin
x
−
cos
x
(
cos
x
+
sin
x
)
n
−
1
−
2
(
n
−
2
)
∫
d
x
(
cos
x
+
sin
x
)
n
−
2
)
∫
cos
a
x
d
x
cos
a
x
+
sin
a
x
=
x
2
+
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
∫
cos
a
x
d
x
cos
a
x
−
sin
a
x
=
x
2
−
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
∫
sin
a
x
d
x
cos
a
x
+
sin
a
x
=
x
2
−
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
∫
sin
a
x
d
x
cos
a
x
−
sin
a
x
=
−
x
2
−
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
∫
cos
a
x
d
x
sin
a
x
(
1
+
cos
a
x
)
=
−
1
4
a
tan
2
a
x
2
+
1
2
a
ln
|
tan
a
x
2
|
+
C
∫
cos
a
x
d
x
sin
a
x
(
1
+
−
cos
a
x
)
=
−
1
4
a
cot
2
a
x
2
−
1
2
a
ln
|
tan
a
x
2
|
+
C
∫
sin
a
x
d
x
cos
a
x
(
1
+
sin
a
x
)
=
1
4
a
cot
2
(
a
x
2
+
π
4
)
+
1
2
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
∫
sin
a
x
d
x
cos
a
x
(
1
−
sin
a
x
)
=
1
4
a
tan
2
(
a
x
2
+
π
4
)
−
1
2
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
∫
sin
a
x
cos
a
x
d
x
=
1
2
a
sin
2
a
x
+
C
∫
sin
a
1
x
cos
a
2
x
d
x
=
−
cos
(
a
1
+
a
2
)
x
2
(
a
1
+
a
2
)
−
cos
(
a
1
−
a
2
)
x
2
(
a
1
−
a
2
)
+
C
(for
|
a
1
|
≠
|
a
2
|
)
∫
sin
n
a
x
cos
a
x
d
x
=
1
a
(
n
+
1
)
sin
n
+
1
a
x
+
C
(for
n
≠
−
1
)
∫
sin
a
x
cos
n
a
x
d
x
=
−
1
a
(
n
+
1
)
cos
n
+
1
a
x
+
C
(for
n
≠
−
1
)
∫
sin
n
a
x
cos
m
a
x
d
x
=
−
sin
n
−
1
a
x
cos
m
+
1
a
x
a
(
n
+
m
)
+
n
−
1
n
+
m
∫
sin
n
−
2
a
x
cos
m
a
x
d
x
(for
m
,
n
>
0
)
also:
∫
sin
n
a
x
cos
m
a
x
d
x
=
sin
n
+
1
a
x
cos
m
−
1
a
x
a
(
n
+
m
)
+
m
−
1
n
+
m
∫
sin
n
a
x
cos
m
−
2
a
x
d
x
(for
m
,
n
>
0
)
∫
d
x
sin
a
x
cos
a
x
=
1
a
ln
|
tan
a
x
|
+
C
∫
d
x
sin
a
x
cos
n
a
x
=
1
a
(
n
−
1
)
cos
n
−
1
a
x
+
∫
d
x
sin
a
x
cos
n
−
2
a
x
(for
n
≠
1
)
∫
d
x
sin
n
a
x
cos
a
x
=
−
1
a
(
n
−
1
)
sin
n
−
1
a
x
+
∫
d
x
sin
n
−
2
a
x
cos
a
x
(for
n
≠
1
)
∫
sin
a
x
d
x
cos
n
a
x
=
1
a
(
n
−
1
)
cos
n
−
1
a
x
+
C
(for
n
≠
1
)
∫
sin
2
a
x
d
x
cos
a
x
=
−
1
a
sin
a
x
+
1
a
ln
|
tan
(
π
4
+
a
x
2
)
|
+
C
∫
sin
2
a
x
d
x
cos
n
a
x
=
sin
a
x
a
(
n
−
1
)
cos
n
−
1
a
x
−
1
n
−
1
∫
d
x
cos
n
−
2
a
x
(for
n
≠
1
)
∫
sin
n
a
x
d
x
cos
a
x
=
−
sin
n
−
1
a
x
a
(
n
−
1
)
+
∫
sin
n
−
2
a
x
d
x
cos
a
x
(for
n
≠
1
)
∫
sin
n
a
x
d
x
cos
m
a
x
=
sin
n
+
1
a
x
a
(
m
−
1
)
cos
m
−
1
a
x
−
n
−
m
+
2
m
−
1
∫
sin
n
a
x
d
x
cos
m
−
2
a
x
(for
m
≠
1
)
also:
∫
sin
n
a
x
d
x
cos
m
a
x
=
−
sin
n
−
1
a
x
a
(
n
−
m
)
cos
m
−
1
a
x
+
n
−
1
n
−
m
∫
sin
n
−
2
a
x
d
x
cos
m
a
x
(for
m
≠
n
)
also:
∫
sin
n
a
x
d
x
cos
m
a
x
=
sin
n
−
1
a
x
a
(
m
−
1
)
cos
m
−
1
a
x
−
n
−
1
m
−
1
∫
sin
n
−
2
a
x
d
x
cos
m
−
2
a
x
(for
m
≠
1
)
∫
cos
a
x
d
x
sin
n
a
x
=
−
1
a
(
n
−
1
)
sin
n
−
1
a
x
+
C
(for
n
≠
1
)
∫
cos
2
a
x
d
x
sin
a
x
=
1
a
(
cos
a
x
+
ln
|
tan
a
x
2
|
)
+
C
∫
cos
2
a
x
d
x
sin
n
a
x
=
−
1
n
−
1
(
cos
a
x
a
sin
n
−
1
a
x
)
+
∫
d
x
sin
n
−
2
a
x
)
(for
n
≠
1
)
∫
cos
n
a
x
d
x
sin
m
a
x
=
−
cos
n
+
1
a
x
a
(
m
−
1
)
sin
m
−
1
a
x
−
n
−
m
−
2
m
−
1
∫
cos
n
a
x
d
x
sin
m
−
2
a
x
(for
m
≠
1
)
also:
∫
cos
n
a
x
d
x
sin
m
a
x
=
cos
n
−
1
a
x
a
(
n
−
m
)
sin
m
−
1
a
x
+
n
−
1
n
−
m
∫
cos
n
−
2
a
x
d
x
sin
m
a
x
(for
m
≠
n
)
also:
∫
cos
n
a
x
d
x
sin
m
a
x
=
−
cos
n
−
1
a
x
a
(
m
−
1
)
sin
m
−
1
a
x
−
n
−
1
m
−
1
∫
cos
n
−
2
a
x
d
x
sin
m
−
2
a
x
(for
m
≠
1
)
∫
sin
a
x
tan
a
x
d
x
=
1
a
(
ln
|
sec
a
x
+
tan
a
x
|
−
sin
a
x
)
+
C
∫
tan
n
a
x
d
x
sin
2
a
x
=
1
a
(
n
−
1
)
tan
n
−
1
(
a
x
)
+
C
(for
n
≠
1
)
∫
tan
n
a
x
d
x
cos
2
a
x
=
1
a
(
n
+
1
)
tan
n
+
1
a
x
+
C
(for
n
≠
−
1
)
∫
cot
n
a
x
d
x
sin
2
a
x
=
1
a
(
n
+
1
)
cot
n
+
1
a
x
+
C
(for
n
≠
−
1
)
∫
cot
n
a
x
d
x
cos
2
a
x
=
1
a
(
1
−
n
)
tan
1
−
n
a
x
+
C
(for
n
≠
1
)
∫
−
c
c
sin
x
d
x
=
0
∫
−
c
c
cos
x
d
x
=
2
∫
0
c
cos
x
d
x
=
2
∫
−
c
0
cos
x
d
x
=
2
sin
c
∫
−
c
c
tan
x
d
x
=
0
∫
−
a
2
a
2
x
2
cos
2
n
π
x
a
d
x
=
a
3
(
n
2
π
2
−
6
)
2
4
n
2
π
2
(for
n
=
1
,
3
,
5
.
.
.
)
∫
arcsin
x
c
d
x
=
x
arcsin
x
c
+
c
2
−
x
2
∫
x
arcsin
x
c
d
x
=
(
x
2
2
−
c
2
4
)
arcsin
x
c
+
x
4
c
2
−
x
2
∫
x
2
arcsin
x
c
d
x
=
x
3
3
arcsin
x
c
+
x
2
+
2
c
2
9
c
2
−
x
2
∫
x
n
sin
−
1
x
d
x
=
1
n
+
1
(
x
n
+
1
sin
−
1
x
+
x
n
1
−
x
2
−
n
x
n
−
1
sin
−
1
x
n
−
1
+
n
∫
x
n
−
2
sin
−
1
x
d
x
)
∫
arccos
x
c
d
x
=
x
arccos
x
c
−
c
2
−
x
2
∫
x
arccos
x
c
d
x
=
(
x
2
2
−
c
2
4
)
arccos
x
c
−
x
4
c
2
−
x
2
∫
x
2
arccos
x
c
d
x
=
x
3
3
arccos
x
c
−
x
2
+
2
c
2
9
c
2
−
x
2
∫
arctan
x
c
d
x
=
x
arctan
x
c
−
c
2
ln
(
c
2
+
x
2
)
∫
x
arctan
x
c
d
x
=
c
2
+
x
2
2
arctan
x
c
−
c
x
2
∫
x
2
arctan
x
c
d
x
=
x
3
3
arctan
x
c
−
c
x
2
6
+
c
3
6
ln
c
2
+
x
2
∫
x
n
arctan
x
c
d
x
=
x
n
+
1
n
+
1
arctan
x
c
−
c
n
+
1
∫
x
n
+
1
d
x
c
2
+
x
2
(
n
≠
1
)
∫
arcsec
x
c
d
x
=
x
arcsec
x
c
+
x
c
|
x
|
ln
|
x
±
x
2
−
1
|
∫
x
arcsec
x
d
x
=
1
2
(
x
2
arcsec
x
−
x
2
−
1
)
∫
x
n
arcsec
x
d
x
=
1
n
+
1
(
x
n
+
1
arcsec
x
−
1
n
(
x
n
−
1
x
2
−
1
+
(
1
−
n
)
(
x
n
−
1
arcsec
x
+
(
1
−
n
)
∫
x
n
−
2
arcsec
x
d
x
)
)
)
∫
a
r
c
c
o
t
x
c
d
x
=
x
a
r
c
c
o
t
x
c
+
c
2
ln
(
c
2
+
x
2
)
∫
x
a
r
c
c
o
t
x
c
d
x
=
c
2
+
x
2
2
a
r
c
c
o
t
x
c
+
c
x
2
∫
x
2
a
r
c
c
o
t
x
c
d
x
=
x
3
3
a
r
c
c
o
t
x
c
+
c
x
2
6
−
c
3
6
ln
(
c
2
+
x
2
)
∫
x
n
a
r
c
c
o
t
x
c
d
x
=
x
n
+
1
n
+
1
a
r
c
c
o
t
x
c
+
c
n
+
1
∫
x
n
+
1
d
x
c
2
+
x
2
(
n
≠
1
)
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
(
n
≠
−
1
)
∫
d
x
a
x
+
b
=
1
a
ln
|
a
x
+
b
|
∫
x
(
a
x
+
b
)
n
d
x
=
a
(
n
+
1
)
x
−
b
a
2
(
n
+
1
)
(
n
+
2
)
(
a
x
+
b
)
n
+
1
(
n
∉
{
1
,
2
}
)
∫
x
a
x
+
b
d
x
=
x
a
−
b
a
2
ln
|
a
x
+
b
|
∫
x
(
a
x
+
b
)
2
d
x
=
b
a
2
(
a
x
+
b
)
+
1
a
2
ln
|
a
x
+
b
|
∫
x
(
a
x
+
b
)
n
d
x
=
a
(
1
−
n
)
x
−
b
a
2
(
n
−
1
)
(
n
−
2
)
(
a
x
+
b
)
n
−
1
(
n
∉
{
1
,
2
}
)
∫
x
2
a
x
+
b
d
x
=
1
a
3
(
(
a
x
+
b
)
2
2
−
2
b
(
a
x
+
b
)
+
b
2
ln
|
a
x
+
b
|
)
∫
x
2
(
a
x
+
b
)
2
d
x
=
1
a
3
(
a
x
+
b
−
2
b
ln
|
a
x
+
b
|
−
b
2
a
x
+
b
)
∫
x
2
(
a
x
+
b
)
3
d
x
=
1
a
3
(
ln
|
a
x
+
b
|
+
2
b
a
x
+
b
−
b
2
2
(
a
x
+
b
)
2
)
∫
x
2
(
a
x
+
b
)
n
d
x
=
1
a
3
(
−
1
(
n
−
3
)
(
a
x
+
b
)
n
−
3
+
2
b
(
n
−
2
)
(
a
+
b
)
n
−
2
−
b
2
(
n
−
1
)
(
a
x
+
b
)
n
−
1
)
(
n
∉
{
1
,
2
,
3
}
)
∫
d
x
x
(
a
x
+
b
)
=
−
1
b
ln
|
a
x
+
b
x
|
∫
d
x
x
2
(
a
x
+
b
)
=
−
1
b
x
+
a
b
2
ln
|
a
x
+
b
x
|
∫
d
x
x
2
(
a
x
+
b
)
2
=
−
a
(
1
b
2
(
a
x
+
b
)
+
1
a
b
2
x
−
2
b
3
ln
|
a
x
+
b
x
|
)
∫
d
x
x
2
+
a
2
=
1
a
arctan
x
a
∫
d
x
x
2
−
a
2
=
−
1
a
a
r
c
t
a
n
h
x
a
=
1
2
a
ln
a
−
x
a
+
x
(
|
x
|
<
|
a
|
)
∫
d
x
x
2
−
a
2
=
−
1
a
a
r
c
c
o
t
h
x
a
=
1
2
a
ln
x
−
a
x
+
a
(
|
x
|
>
|
a
|
)
∫
d
x
a
x
2
+
b
x
+
c
=
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
(
4
a
c
−
b
2
>
0
)
∫
d
x
a
x
2
+
b
x
+
c
=
2
b
2
−
4
a
c
a
r
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
=
1
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
(
4
a
c
−
b
2
<
0
)
∫
d
x
a
x
2
+
b
x
+
c
=
−
2
2
a
x
+
b
(
4
a
c
−
b
2
=
0
)
∫
x
a
x
2
+
b
x
+
c
d
x
=
1
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
b
2
a
∫
d
x
a
x
2
+
b
x
+
c
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
+
2
a
n
−
b
m
a
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
(
4
a
c
−
b
2
>
0
)
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
+
2
a
n
−
b
m
a
b
2
−
4
a
c
a
r
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
(
4
a
c
−
b
2
<
0
)
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
2
a
n
−
b
m
a
(
2
a
x
+
b
)
(
4
a
c
−
b
2
=
0
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
=
2
a
x
+
b
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
+
(
2
n
−
3
)
2
a
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
∫
x
(
a
x
2
+
b
x
+
c
)
n
d
x
=
b
x
+
2
c
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
−
b
(
2
n
−
3
)
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
∫
d
x
x
(
a
x
2
+
b
x
+
c
)
=
1
2
c
ln
|
x
2
a
x
2
+
b
x
+
c
|
−
b
2
c
∫
d
x
a
x
2
+
b
x
+
c
∫
r
d
x
=
1
2
(
x
r
+
a
2
ln
(
x
+
r
)
)
∫
r
3
d
x
=
1
4
x
r
3
+
3
8
a
2
x
r
+
3
8
a
4
ln
(
x
+
r
)
∫
r
5
d
x
=
1
6
x
r
5
+
5
2
4
a
2
x
r
3
+
5
1
6
a
4
x
r
+
5
1
6
a
6
ln
(
x
+
r
)
∫
x
r
d
x
=
r
3
3
∫
x
r
3
d
x
=
r
5
5
∫
x
r
2
n
+
1
d
x
=
r
2
n
+
3
2
n
+
3
∫
x
2
r
d
x
=
x
r
3
4
−
a
2
x
r
8
−
a
4
8
ln
(
x
+
r
)
∫
x
2
r
3
d
x
=
x
r
5
6
−
a
2
x
r
3
2
4
−
a
4
x
r
1
6
−
a
6
1
6
ln
(
x
+
r
)
∫
x
3
r
d
x
=
r
5
5
−
a
2
r
3
3
∫
x
3
r
3
d
x
=
r
7
7
−
a
2
r
5
5
∫
x
3
r
2
n
+
1
d
x
=
r
2
n
+
5
2
n
+
5
−
a
3
r
2
n
+
3
2
n
+
3
∫
x
4
r
d
x
=
x
3
r
3
6
−
a
2
x
r
3
8
+
a
4
x
r
1
6
+
a
6
1
6
ln
(
x
+
r
)
∫
x
4
r
3
d
x
=
x
3
r
5
8
−
a
2
x
r
5
1
6
+
a
4
x
r
3
6
4
+
3
a
6
x
r
1
2
8
+
3
a
8
1
2
8
ln
(
x
+
r
)
∫
x
5
r
d
x
=
r
7
7
−
2
a
2
r
5
5
+
a
4
r
3
3
∫
x
5
r
3
d
x
=
r
9
9
−
2
a
2
r
7
7
+
a
4
r
5
5
∫
x
5
r
2
n
+
1
d
x
=
r
2
n
+
7
2
n
+
7
−
2
a
2
r
2
n
+
5
2
n
+
5
+
a
4
r
2
n
+
3
2
n
+
3
∫
r
d
x
x
=
r
−
a
ln
|
a
+
r
x
|
=
r
−
a
arsinh
a
x
∫
r
3
d
x
x
=
r
3
3
+
a
2
r
−
a
3
ln
|
a
+
r
x
|
∫
r
5
d
x
x
=
r
5
5
+
a
2
r
3
3
+
a
4
r
−
a
5
ln
|
a
+
r
x
|
∫
r
7
d
x
x
=
r
7
7
+
a
2
r
5
5
+
a
4
r
3
3
+
a
6
r
−
a
7
ln
|
a
+
r
x
|
∫
d
x
r
=
arsinh
x
a
=
ln
(
x
+
r
a
)
∫
d
x
r
3
=
x
a
2
r
∫
x
d
x
r
=
r
∫
x
d
x
r
3
=
−
1
r
∫
x
2
d
x
r
=
x
2
r
−
a
2
2
arsinh
x
a
=
x
2
r
−
a
2
2
ln
(
x
+
r
a
)
∫
d
x
x
r
=
−
1
a
arsinh
a
x
=
−
1
a
ln
|
a
+
r
x
|
Categories
:
Tích phân
VI
Navigation menu
Personal tools
Log in
Namespaces
Page
Discussion
beta
Views
Read
View source
View history
More
Search
Navigation
Main page
Recent changes
Random page
Help about MediaWiki
Special pages
Tools
What links here
Related changes
Printable version
Permanent link
Page information
Cite this page