Begin2.DVI

∫π

0

sinpθsinqθ dθ=

{ 0 , p 6 =q

π

2 , p=q

∫π

0

sinpθcosqθ dθ=







0 , p+qeven

2 p

p^2 −q^2 , p+qodd

∫π

0

x dx

a^2 −cos^2 x=

π^2

2 a

√

a^2 − 1

∫π

0

dx
a+bcosx=
√ π
a^2 −b^2
46.

∫π

0

sinθ dθ

1 − 2 acosθ+a^2 =

2

atanh

− (^1) a
47.
∫π
0
sin 2θ dθ
1 − 2 acosθ+a^2 =
2
a^2 (1 +a
(^2) ) tanh−^1 a−^2
a
48.
∫π
0
xsinx dx
1 − 2 acosx+a^2 =



π
aln(1 +a), |a|<^1
πln
(
1 +^1 a
)
, |a|> 1
49.
∫π
0
cospθ dθ
1 − 2 acosθ+a^2 =



πap
1 −a^2 , a
(^2) < 1
πa−p
a^2 − 1 , a
(^2) > 1
50.
∫π
0
cospθ dθ
(1− 2 acosθ+a^2 )^2 =



πap
(1−a^2 )^3 [(p+ 1)−(p−1)a
(^2) ], a (^2) < 1
πa−p
(a^2 −1)^3 [(1−p) + (1 +p)a
(^2) ], a (^2) > 1
51.
∫π
0
cospθ dθ
(1− 2 acosθ+a^2 )^3 =



πap
2(1−a^2 )^5
[
(p+ 2)(p+ 1) + 2(p+ 2)(p−2)a^2 + (p−2)(p−1)a^4
]
, a^2 < 1
πa−p
2(a^2 −1)^5
[
(1−p)(2−p) + 2(2−p)(2 +p)a^2 + (2 +p)(1 +p)a^4
]
, a^2 > 1
52.
∫ 2 π
0
dx
(a+bsinx)^2 =
2 πa
(a^2 −b^2 )^3 /^2
53.
∫ 2 π
0
dx
a+bsinx=
√^2 π
a^2 −b^2
54.
∫ 2 π
0
dx
a+bcosx=
√^2 π
a^2 −b^2
55.
∫ 2 π
0
dx
(a+bsinx)^2 =
2 πa
(a^2 −b^2 )^3 /^2
56.
∫ 2 π
0
dx
(a+bcosx)^2 =
2 πa
(a^2 −b^2 )^3 /^2
57.
∫L
−L
sinmπxL sinnπxL dx=
{ 0 , m 6 =n, m, nintegers
L
2 , m=n
Appendix C