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Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/127
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(168)
{
∫
0
ϖ
1
−
r
Cos
.
p
1
−
2
r
Cos
.
p
+
r
2
.
Cos
.
a
2
(
ϖ
−
p
)
(
Sin
.
p
2
)
a
.
d
p
=
2
a
−
1
.
ϖ
[
1
+
(
1
−
r
)
−
a
]
,
∫
0
ϖ
r
Sin
.
p
1
−
2
r
Cos
.
p
+
r
2
.
Sin
.
a
2
(
ϖ
−
p
)
(
Sin
.
p
2
)
a
.
d
p
=
2
a
−
1
.
ϖ
[
1
−
(
1
−
r
)
−
a
]
.
{\displaystyle \left\{{\begin{aligned}&\int _{0}^{\varpi }{\frac {1-r\operatorname {Cos} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.{\frac {\operatorname {Cos} .{\frac {a}{2}}(\varpi -p)}{\left(\operatorname {Sin} .{\frac {p}{2}}\right)^{a}}}.\operatorname {d} p=2^{a-1}.\varpi \left[1+(1-r)^{-a}\right],\\\\&\int _{0}^{\varpi }{\frac {r\operatorname {Sin} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.{\frac {\operatorname {Sin} .{\frac {a}{2}}(\varpi -p)}{\left(\operatorname {Sin} .{\frac {p}{2}}\right)^{a}}}.\operatorname {d} p=2^{a-1}.\varpi \left[1-(1-r)^{-a}\right].\end{aligned}}\right.}
(169)
∫
0
ϖ
1
−
r
Cos
.
p
1
−
2
r
Cos
.
p
+
r
2
.
l
.
Tang
.
p
2
d
p
=
π
2
l
(
1
−
r
1
+
r
)
.
{\displaystyle \int _{0}^{\varpi }{\frac {1-r\operatorname {Cos} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.\operatorname {l} .\operatorname {Tang} .{\frac {p}{2}}\operatorname {d} p={\frac {\pi }{2}}\operatorname {l} \left({\frac {1-r}{1+r}}\right).}
(170)
{
∫
0
ϖ
1
−
r
Cos
.
p
1
−
2
r
Cos
.
p
+
r
2
.
l
Cos
.
p
2
d
p
=
π
2
l
(
1
+
r
4
)
,
∫
0
ϖ
r
Sin
.
p
1
−
2
r
Cos
.
p
+
r
2
p
d
p
=
ϖ
l
(
1
+
r
)
.
{\displaystyle \left\{{\begin{aligned}&\int _{0}^{\varpi }{\frac {1-r\operatorname {Cos} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.\operatorname {l} \operatorname {Cos} .{\frac {p}{2}}\operatorname {d} p={\frac {\pi }{2}}\operatorname {l} \left({\frac {1+r}{4}}\right),\\\\&\int _{0}^{\varpi }{\frac {r\operatorname {Sin} .p}{1-2r\operatorname {Cos} .p+r^{2}}}p\operatorname {d} p=\varpi \operatorname {l} (1+r).\end{aligned}}\right.}
(171)
∫
0
ϖ
1
−
r
Cos
.
p
1
−
2
r
Cos
.
p
+
r
2
.
l
.
Sin
.
p
2
d
p
=
π
2
l
(
1
−
r
4
)
.
{\displaystyle \int _{0}^{\varpi }{\frac {1-r\operatorname {Cos} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.\operatorname {l} .\operatorname {Sin} .{\frac {p}{2}}\operatorname {d} p={\frac {\pi }{2}}\operatorname {l} \left({\frac {1-r}{4}}\right).}
(172)
{
∫
0
ϖ
1
−
r
Cos
.
p
1
−
2
r
Cos
.
p
+
r
2
.
l
.
Cos
.
1
2
p
(
p
2
)
2
+
(
l
.
Cos
.
p
2
)
2
d
p
=
π
1
−
r
−
π
2
[
1
l
(
2
)
+
1
l
(
2
)
−
l
(
1
+
r
)
]
,
∫
0
ϖ
r
Sin
.
p
1
−
2
r
Cos
.
p
+
r
2
.
1
2
p
(
p
2
)
2
+
(
l
.
Cos
.
p
2
)
2
.
d
p
=
−
π
2
[
1
l
(
2
)
−
1
l
(
2
)
−
l
(
1
+
r
)
]
;
{\displaystyle \left\{{\begin{aligned}&\int _{0}^{\varpi }{\frac {1-r\operatorname {Cos} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.{\frac {\operatorname {l} .\operatorname {Cos} .{\frac {1}{2}}p}{\left({\frac {p}{2}}\right)^{2}+\left(\operatorname {l} .\operatorname {Cos} .{\frac {p}{2}}\right)^{2}}}\operatorname {d} p\\&\qquad \qquad \qquad \qquad ={\frac {\pi }{1-r}}-{\frac {\pi }{2}}\left[{\frac {1}{\operatorname {l} (2)}}+{\frac {1}{\operatorname {l} (2)-\operatorname {l} (1+r)}}\right],\\\\&\int _{0}^{\varpi }{\frac {r\operatorname {Sin} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.{\frac {{\frac {1}{2}}p}{\left({\frac {p}{2}}\right)^{2}+\left(\operatorname {l} .\operatorname {Cos} .{\frac {p}{2}}\right)^{2}}}.\operatorname {d} p\\&\qquad \qquad \qquad \qquad =-{\frac {\pi }{2}}\left[{\frac {1}{\operatorname {l} (2)}}-{\frac {1}{\operatorname {l} (2)-\operatorname {l} (1+r)}}\right]\,;\end{aligned}}\right.}
et ainsi du reste.
On trouvera, au contraire, pour
r
2
>
1
,
{\displaystyle r^{2}>1,}
(173)
{
∫
0
ϖ
1
−
r
Cos
.
p
1
−
2
r
Cos
.
p
+
r
2
.
(
Tang
.
p
2
)
a
d
p
=
π
2
Cos
.
a
π
2
[
1
−
(
r
−
1
r
+
1
)
a
]
,
∫
0
ϖ
r
Sin
.
p
1
−
2
r
Cos
.
p
+
r
2
.
(
Tang
.
p
2
)
a
d
p
=
π
2
Sin
.
a
π
2
[
1
−
(
r
−
1
r
+
1
)
a
]
.
{\displaystyle \left\{{\begin{aligned}&\int _{0}^{\varpi }{\frac {1-r\operatorname {Cos} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.\left(\operatorname {Tang} .{\frac {p}{2}}\right)^{a}\operatorname {d} p={\frac {\pi }{2\operatorname {Cos} .{\frac {a\pi }{2}}}}\left[1-\left({\frac {r-1}{r+1}}\right)^{a}\right],\\\\&\int _{0}^{\varpi }{\frac {r\operatorname {Sin} .p}{1-2r\operatorname {Cos} .p+r^{2}}}.\left(\operatorname {Tang} .{\frac {p}{2}}\right)^{a}\operatorname {d} p={\frac {\pi }{2\operatorname {Sin} .{\frac {a\pi }{2}}}}\left[1-\left({\frac {r-1}{r+1}}\right)^{a}\right].\end{aligned}}\right.}