高等数学
积 分 表
公 式 推 导
目
录
（一）含有 ax + b 的积分（1~9）·······················································1
（二）含有 ax + b 的积分（10~18）···················································5
（三）含有 x 2 ± a 2 的积分（19~21）····················································9
2
（四）含有 ax + b (a > 0) 的积分（22~28）············································11
（五）含有 ax 2 + bx + c (a > 0) 的积分（29~30）········································14
（六）含有
x 2 + a 2 ( a > 0 ) 的积分（31~44）·········································15
（七）含有 x 2 − a 2 (a > 0) 的积分（45~58）·········································24
（八）含有 a 2 − x 2 ( a > 0 ) 的积分（59~72）·········································37
（九）含有 ± a 2 + bx + c ( a > 0) 的积分（73~78）····································48
（十）含有
±
x − a
x − b
或
( x − a )( b − x )
的积分（79~82）···························51
（十一）含有三角函数的积分（83~112）···········································55
（十二）含有反三角函数的积分（其中 a > 0)（113~121）·······················68
（十三）含有指数函数的积分（122~131）··········································73
（十四）含有对数函数的积分（132~136）··········································78
（十五）含有双曲函数的积分（137~141）··········································80
（十六）定积分 （142~147）····························································81
常数
和基
本初等
函数
导数
公式·········································85
附录：
录：常数
常数和基
和基本
初等函数
函数导数
导数公
（一）含有 ax + b 的积分（1~9）
1.
dx
1
∫ ax + b = a ⋅ ln
ax + b + C
1
b
的定义域为{ x | x ≠ − }
ax + b
a
1
令 ax + b = t (t ≠ 0) ，则dt = adx ,∴ dx = dt
a
dx
1 1
∴∫
= ∫ dt
ax + b a t
1
= ⋅ ln t + C
a
dx
1
将 t = ax + b 代入上式得：
∫ ax + b = a ⋅ ln ax + b + C
证明：
被积函数 f ( x ) =
2.
∫ (ax + b)
μ
1
⋅ ( ax + b ) μ +1 + C
a ( μ + 1)
dx =
证明：
令 ax + b = t , 则dt = adx ,∴ dx =
( μ ≠ −1)
1
dt
a
1 μ
t dt
a∫
1
=
⋅ t μ +1 + C
a ( μ + 1)
∴ ∫ ( ax + b) μ dx =
μ
将 t = ax + b代入上式得：
∫ (ax + b) dx =
3.
x
1
∫ ax + b dx = a (ax + b − b⋅ ln
2
ax + b
1
⋅ ( ax + b) μ +1 + C
a ( μ + 1)
)+ C
x
b
的定义域为 { x | x ≠ − }
ax + b
a
1
1
令 ax + b = t (t ≠ 0) , 则 x = (t − b ) , dx = dt
a
a
1
(t − b ) 1
x
1 ⎛
b⎞
∴∫
dx = ∫ a
· dt = 2 ∫ ⎜ 1 − ⎟ dt
ax + b
t
a
a ⎝
t⎠
1
1 b
= 2 ∫ dt − 2 ∫ dt
a
a
t
t
b
= 2 − 2 ⋅ ln t + C
a
a
1
= 2 (t − b⋅ ln t ) + C
a
x
1
将 t = ax + b 代入上式得：∫
dx = 2 (ax + b − b⋅ ln ax + b
ax + b
a
证明：被积函数 f ( x ) =
)+ C
-1-
4.
5.
x2
1 ⎡1
⎤
2
2
∫ ax + bdx = a 3 ⎢⎣ 2 (ax + b) − 2b ( ax + b) + b ⋅ ln ax + b ⎥⎦ + C
x2
1 (ax + b) 2 − 2abx − b 2 )
证明：∫
dx = 2 ∫
dx
ax + b
ax + b
a
1
1 2abx
1
b2
= 2 ∫ (ax + b)dx − 2 ∫
dx − 2 ∫
dx
a
a ax + b
a ax + b
1
1
∵ 2 ∫ (ax + b)dx = 3 (ax + b) 2 + C1
a
2a
1 2abx
2b ax + b − b
dx = 3 ∫
d ( ax)
ax + b
a 2 ∫ ax + b
a
2b
2b 2
1
= 3 ∫ dx − 3 ∫
d (ax + b)
a
a
ax + b
2b
2b 2
= 3 x − 3 ln ax + b + C 2
a
a
2
2
1
b
b
1
b2
dx
=
d
(
ax
+
b
)
=
ln ax + b + C 3
a 2 ∫ ax + b
a 3 ∫ ax + b
a3
x2
1 ⎡1
⎤
2
2
由以上各式整理得：
∫ ax + bdx = a 3 ⎢⎣ 2 (ax + b) − 2b (ax + b) + b ⋅ ln ax + b ⎥⎦ + C
dx
ax + b
+C
x
1
b
证明：被积函数 f ( x ) =
的定义域为 { x | x ≠ − }
x ⋅ (ax + b)
a
1
A
B
设
= +
, 则 1 = A( ax + b ) + Bx = (Aa + B) x + Ab
x ⋅ (ax + b ) x ax + b
1
∫ x (ax + b) = − b ⋅ ln
1
⎧
A=
⎪
⎧ Aa + B = 0
⎪
b
∴ 有 ⎨
⇒ ⎨
⎩ Ab = 1
⎪B = − a
⎪⎩
b
dx
1
a
1 1
a
1
于是 ∫
= ∫[ −
]dx = ∫ dx − ∫
dx
x (ax + b)
bx b ⋅ ( ax + b )
b x
b ax + b
1 1
1
1
= ∫ dx − ∫
d ( ax + b )
b x
b ax + b
1
1
= ⋅ ln x − ⋅ ln ax + b + C
b
b
1
x
= ⋅ ln
+C
提示：
log a b −1 = − log a b
b
ax + b
1
ax + b
= − ⋅ ln
+C
b
x
-2-
6.
dx
1
a
ax + b
= − + 2 ⋅ ln
+C
bx b
x
( ax + b )
1
b
证明：被积函数 f ( x ) = 2
的定义域为 { x | x ≠ − }
a
x ⋅ (ax + b)
1
A B
C
设 2
= + 2 +
, 则 1 = Ax ( ax + b) + B( ax + b) + Cx 2
ax + b
x ⋅ (ax + b) x x
∫x
2
即 x 2 (Aa + C) + x ( Ab + aB ) + Bb = 1
a
⎧
⎪A = − b2
⎧ Aa + C = 0
⎪
1
⎪
⎪
∴ 有 ⎨ Ab + aB = 0 ⇒ ⎨ B =
b
⎪ Bb = 1
⎪
⎩
⎪
a2
C
=
⎪
b2
⎩
dx
a 1
1 1
a2
于是 ∫ 2
= − 2 ∫ dx + ∫ 2 dx + 2
x (ax + b)
b x
b x
b
1
∫ ax + b dx
a 1
1 1
a
1
dx + ∫ 2 dx + 2 ∫
d ( ax + b )
2 ∫
b x
b x
b ax + b
a
1
a
= − 2 ⋅ ln x −
+ 2 ⋅ ln ax + b + C
bx b
b
1
a
ax + b
=−
+ 2 ⋅ ln
+C
bx b
x
=−
7.
x
1 ⎛
b ⎞
dx = 2 ⎜ ln ax + b +
⎟+C
2
ax + b ⎠
( ax + b)
a ⎝
x
b
证明：被积函数 f ( x ) =
的定义域为 { x | x ≠ − }
2
a
( ax + b )
x
A
B
设
=
+
, 则 x = A( ax + b ) + B
2
ax + b ( ax + b ) 2
( ax + b)
即 x ⋅ Aa + ( Ab + B) = x
∫
⎧ Aa = 1
∴ 有 ⎨
⎩ Ab + B = 0
于是
∫
⇒
1
⎧
⎪⎪ A = a
⎨
⎪B = − b
⎩⎪
a
x
1
1
b
1
dx = ∫
dx − ∫
dx
2
a ax + b
a (ax + b ) 2
( ax + b)
1
1
b
1
= 2 ∫
d ( ax + b ) − 2 ∫
d (ax + b)
a ax + b
a ( ax + b ) 2
1
b
= 2 ⋅ ln ax + b + 2
+C
a
a (ax + b)
=
1 ⎛
b ⎞
ln ax + b +
⎟+C
2 ⎜
ax + b ⎠
a ⎝
-3-
8.
x2
1
∫ ( ax + b) 2 dx = a 3
⎛
b2 ⎞
⎜⎜ ax + b − 2 b⋅ ln ax + b −
⎟+C
ax + b ⎟⎠
⎝
x2
b
证明：被积函数 f ( x ) =
的定义域为{x | x ≠ − }
2
(ax + b)
a
1
1
令 ax + b = t (t ≠ 0) , 则 x = (t − b ) , dx = dt
a
a
2
2
2
2
x
(b − t )
b + t − 2bt
∴
=
=
2
2 2
(ax + b)
a t
a 2t 2
∴
x2
b 2 + t 2 − 2bt
b2 1
1
2b 1
dx
=
dt
=
dt + 3 ∫ dt − 3 ∫ dt
3 ∫ 2
∫ (ax + b) 2 ∫ a 3t 2
a t
a
a t
b2
1
2b
= − 3 + 3 ⋅ t − 3 ⋅ ln t + C
a t a
a
1
b2
= 3 (t − 2b ⋅ ln t − ) + C
a
t
2
x
1 ⎛
b2
⎜
将 t = ax + b 代入上式得：
dx
=
ax
+
b
−
2
b
⋅
ln
ax
+
b
−
∫ ( ax + b) 2
ax + b
a 3 ⎜⎝
9.
dx
1
1
ax + b
− 2 ·ln |
|+C
b ( ax + b ) b
x
1
b
证明：被积函数 f ( x ) =
的定义域为 { x | x ≠ − }
2
a
x ( ax + b )
1
A
B
D
设：
= +
+
2
x ax + b ( ax + b ) 2
x ( ax + b )
∫ x ( ax + b )
2
=
则 1 = A ( ax + b ) 2 + Bx ( ax + b ) + Dx
= Aa 2 x 2 + Ab 2 + 2 Aabx + Bax 2 + Bbx + Dx
= x 2 ( Aa 2 + Ba ) + x ( 2 Aab + Bb + D ) + Ab 2
1
⎧
A= 2
⎪
b
⎧ Aa 2 + Ba = 0
⎪
a
⎪
⎪
∴ 有 ⎨ 2 Aab + Bb + D = 0 ⇒ ⎨ B = − 2
b
⎪ Ab 2 = 1
⎪
⎩
⎪
a
⎪D = − b
⎩
dx
1 1
a
1
a
1
于是 ∫
= 2 ∫ dx − 2 ∫
dx − ∫
dx
x ( ax + b ) b
x
b
ax + b
b ( ax + b ) 2
1
1
1
1
= 2 ⋅ ln |x| − 2 ⋅ ln |ax + b| + ·
+C
b
b
b ax + b
1
1
ax + b
=
− 2 · ln |
|+ C
b ( ax + b ) b
x
-4-
⎞
⎟⎟ + C
⎠
（二）含有 ax + b 的积分（10~18）
10.
∫
ax + b dx =
2
⋅ ( ax + b) 3 + C
3a
1
1
+1
1
1 1
2
2
证明：
ax
+
b
dx
=
(
ax
+
b
)
d
(
ax
+
b
)
=
⋅
⋅
(
ax
+
b
)
+C
∫
1
a∫
a
1+
2
2
=
⋅ ( ax + b) 3 + C
3a
11.
12.
2
⋅ (3ax − 2b ) ⋅ ( ax + b ) 3 + C
2
15a
t2 −b
2t
t2 −b
证明：令 ax + b = t (t ≥ 0) , 则x =
, dx = dt , x ax + b =
⋅t
a
a
a
t2 −b
2t
2
∴ ∫ x ax + b dx = ∫
⋅ t ⋅ dt = 2 ∫ (t 4 − bt 2 ) dt
a
a
a
2
2b
2
2b
= 2 ∫ dt 5 − 2 ∫ dt 3 = 2 ⋅ t 5 − 2 ⋅ t 3 + C
5a
3a
5a
3a
3
2t
=
(3t 2 − 5b ) + C
2
15a
2
3
将t = ax + b代入上式得：
∫ x ax + b dx = 15a 2 [3(ax + b) − 5b] ⋅ (ax + b) + C
2
=
⋅ (3ax − 2b) ⋅ ( ax + b) 3 + C
15a 2
∫x
ax + b dx =
2
⋅ (15a 2 x 2 − 12abx + 8b 2 ) ⋅ ( ax + b) 3 + C
3
105a
t2 −b
2t
证明：令 ax + b = t (t ≥ 0) , 则x =
, dx = dt ,
a
a
2
2
5
2
3
(t − b )
t + b t − 2bt
x 2 ax + b =
⋅t =
2
a
a2
2
∴ ∫ x 2 ax + b dx = 3 ∫ t ⋅ (t 5 + b 2 t − 2bt 3 )dt
a
2
2b 2
4b
= 3 ∫ t 6 dt − 3 ∫ t 2 dt − 3 ∫ t 4 dt
a
a
a
2
2
1
2b
1
4b 1
= 3⋅
⋅ t 6 +1 + 3 ⋅
⋅ t 1+ 2 − 3 ⋅
⋅ t 4 +1 + C
a 1+ 6
a 1+ 2
a 1+ 4
2
2
2b
4b
= 3 ⋅t7 + 3 ⋅t3 − 3 ⋅t5 + C
7a
3a
5a
3
2t
=
⋅ (15t 4 + 35b 2 − 42bt 2 ) + C
3
105a
将t = ax + b代入上式得：
∫x
2
ax + b dx =
∫x
2
2
⋅ ( ax + b ) 3 15a 2 x 2 + 15b 2 + 30abx + 35b 2 − 42b ⋅ ( ax + b)
3
105a
2
=
⋅ (15a 2 x 2 − 12abx + 8b 2 ) ⋅ ( ax + b) 3 + C
3
105a
ax + b dx =
[
]
-5-
13.
∫
x
ax + b
dx =
2
⋅ ( ax − 2b ) ⋅ ( ax + b ) + C
3a 2
证明：令 ax + b = t (t > 0) , 则 x =
t2 −b
2t
, dx = dt ,
a
a
x
t 2 − b 2t
dx
=
∫ ax + b
∫ at ⋅ a dt
2
2
= 2 ∫ t 2 dt − 2 ∫ bdt
a
a
2
1
2b
= 2⋅
⋅ t 2 +1 − 2 ⋅ t + C
a 1+ 2
a
2
2b
= 2 ⋅t3 − 2 ⋅t + C
3a
a
x
2
2b
将t = ax + b代入上式得：
∫ ax + b dx = 3a 2 ⋅ (ax + b) ⋅ (ax + b) − a 2 ⋅ (ax + b) + C
2
= 2 ⋅ ( ax − 2b) ⋅ ( ax + b) + C
3a
∴
x2
2
14. ∫
dx =
⋅ (3a 2 x 2 − 4abx + 8b 2 ) ⋅ ( ax + b) + C
3
15a
ax + b
t2 − b
2t
证明：令 ax + b = t (t > 0) , 则 x =
, dx = dt ,
a
a
2
2
x
t − b 2 1 2t
∴∫
dx = ∫ (
) ⋅ ⋅ dt
a
t a
ax + b
2
= 3 ∫ (t 4 + b 2 − 2bt 2 )dt
a
2
2
4b
= 3 ∫ t 4 dt + 3 ∫ b 2 dt − 3 ∫ t 2 dt
a
a
a
2 1
2b
= 3 ( t 5 + b2t − t 3 ) + C
3
a 5
2t
=
⋅ (3t 4 + 15b 2 − 10bt 2 ) + C
3
15a
将 t = ax + b代入上式得：
∫
-6-
x2
ax + b
2
⋅ ( ax + b ) ⋅ 3( a 2 x 2 + b 2 + 2abx) + 15b 2 − 10b ⋅ ( ax + b) ⋅ ( ax + b) + C
3
15a
2
=
⋅ (3a 2 x 2 − 4abx + 8b 2 ) ⋅ ( ax + b) + C
3
15a
dx =
[
]
15.
∫
⎧
⎪
dx
⎪
=⎨
x ax + b ⎪
⎪
⎩
1
b
ax + b − b
+C
ax + b + b
⋅ ln
2
−b
⋅ arctan
(b > 0 )
ax + b
+C
−b
证明：令 ax + b = t (t > 0) , 则 x =
(b < 0)
t2 −b
2t
, dx = dt ,
a
a
dx
1
2t
=∫ 2
⋅ dt
a
t −b
ax + b
⋅t
a
2
=∫ 2
dt
t −b
2
1
1.当 b > 0 时 ，∫ 2
dt = 2 ∫ 2
dt
t −b
t − ( b)2
∴
∫x
=
1
b
⋅ ln
将 t = ax + b代入上式得：
∫
2.当 b < 0 时 ，∫
t− b
t+ b
公式 21 ：∫
dx
1
=
⋅ ln
x ax + b
b
2
1
dt = 2 ∫ 2
dt
t −b
t + ( − b)2
2
t
=
⋅ arctan
+C
−b
−b
dx
x−a
+C
x+ a
+C
2
将 t = ax + b代入上式得：
∫
dx
1
=
⋅ ln
x2 − a2
2a
=
2
ax + b − b
+C
ax + b + b
公式 19 ：∫
⋅ arctan
dx
1
x
= ⋅ arctan
+C
x2 + a2
a
a
ax + b
+C
−b
x ax + b
−b
⎧ 1
ax + b − b
⋅ ln
+C
⎪
ax + b + b
dx
⎪ b
综合讨论 1 , 2 得：∫
=⎨
x ax + b ⎪ 2
ax + b
⋅ arctan
+C
⎪
−b
⎩ −b
(b > 0 )
(b > 0 )
-7-
16 .
∫x
2
dx
ax + b a
dx
=−
−
∫
bx
2b x ax + b
ax + b
证明：设
1
2
x ⋅ ax + b
=
A
x ax + b
⎧ A + Ba = 0
∴ 有 ⎨
⎩ Bb = 1
于是
∫x
dx
2
⇒
+
B ax + b
, 则 1 = Ax + B( ax + b)
x2
a
⎧
⎪⎪ A = − b
⎨
⎪B = 1
⎪⎩
b
a
1
1
ax + b
dx + ∫
dx
∫
b x ax + b
b
x2
ax + b
a
1
1
1
=− ∫
dx − ∫ ax + b d
b x ax + b
b
x
=−
=−
a
1
ax + b 1 1
dx −
+ ∫ d ax + b
∫
b x ax + b
bx
b x
a
1
=− ∫
dx −
b x ax + b
=−
=−
17.
-8-
1
−
ax + b 1 1 a
+ ∫ ⋅ ( ax + b ) 2 dx
bx
b x 2
a
1
ax + b a
1
dx −
+
dx
∫
∫
b x ax + b
bx
2b x ax + b
ax + b a
dx
−
∫
bx
2b x ax + b
ax + b
dx
dx = 2 ax + b + b ∫
x
x ax + b
t2 −b
2t
证明：令 ax + b = t (t ≥ 0) , 则 x =
, dx = dt
a
a
2
ax + b
at 2t
t
∴ ∫
dx = ∫ 2
⋅ dt = 2 ∫ 2
dt
x
t −b a
t −b
t 2 − b 2 + b2
1
= 2∫
dt = 2 ∫ dt + 2b ∫ 2
dt
2
t −b
t −b
1
= 2t + 2b ∫ 2
dt
t −b
1
∵ b取值为R , 符号可正可负 ∴ ∫ 2
dt 不能明确积分
t −b
ax + b
1
∴∫
dx = 2t + 2b ∫ 2
dt
x
t −b
1
a
= 2t + 2b ∫ 2
⋅ dx
t − b 2t
ax + b
1
a
将t = ax + b代入上式得：
∫ x dx = 2 (ax + b) + 2b ∫ ax + b − b ⋅ 2 ax + b dx
dx
= 2 ax + b + b ∫
x ax + b
∫
18.
∫
ax + b
ax + b a
dx
dx = −
+ ∫
2
x
2 x ax + b
x
证明：∫
ax + b
1
dx = − ∫ ax + b d
2
x
x
ax + b
1
=−
+ ∫ d ax + b
x
x
1
−
ax + b
1
a
=−
+ ∫ ⋅ ( ax + b) 2 ⋅ dx
x
x
2
ax + b a
dx
=−
+ ∫
x
2 x ax + b
（三）含有 x 2 ± a 2 的积分（19~21）
19.
dx
1
x
= ⋅ arctan + C
2
+a
a
a
π
π
证明：令 x = a ⋅ tant ( − < t < ) , 则 dx = d ( a ⋅ tant ) = a ⋅ sec 2 t dt
2
2
1
dx
1
= 2
= 2 2
2
2
2
x +a
a ⋅ (1 + tan t ) a sec t
dx
1
∴ ∫ 2
= ∫ 2 2 ⋅ a ⋅ sec 2 t dt
2
x +a
a sec t
1
= ∫ dt
a
1
= ⋅t + C
a
x
∵ x = a ⋅ tant
∴ t = arctan
a
x
dx
1
x
将t = arctan 代入上式得：
= ⋅ arctan + C
2
2
∫
a
x +a
a
a
∫x
2
-9-
20.
dx
x
2n − 3
dx
=
+
2 n
2
2
2 n −1
2 ∫
2
+a )
2( n − 1) ⋅ a ⋅ ( x + a )
2( n − 1) ⋅ a ( x + a 2 ) n −1
dx
x
1
证明：∫ 2
= 2
−∫xd 2
2 n
2 n
(x + a )
(x + a )
(x + a 2 ) n
x
= 2
− ∫ x ⋅ ( −n) ⋅ ( x 2 + a 2 ) −n −1 ⋅ 2 x dx
2 n
(x + a )
∫ (x
2
x
x2
= 2
+ 2n ∫ 2
dx
(x + a2 )n
( x + a 2 ) n +1
x
x2 + a2 − a2
= 2
+ 2n ∫ 2
dx
(x + a2 )n
( x + a 2 ) n +1
x
1
1
= 2
+ 2n ∫ 2
dx − 2na 2 ∫ 2
dx
2 n
2 n
(x + a )
(x + a )
( x + a 2 ) n +1
dx
x
1
移项并整理得：(1 − 2 n) ∫ 2
= 2
− 2na 2 ∫ 2
dx
2 n
2 n
(x + a )
(x + a )
( x + a 2 ) n +1
∴
∫ (x
令n + 1 = n , 则 ∫
21.
- 10 -
2
1
1
dx =
2 n +1
+a )
2na 2
⎡
⎤
x
dx
+ ( 2n − 1) ∫ 2
⎢ 2
2 n
2 n ⎥
(x + a ) ⎦
⎣ (x + a )
⎡
⎤
dx
1
x
dx
=
+ ( 2n − 3) ∫ 2
2 n
2 ⎢
2
2 n −1
2 n −1 ⎥
(x + a )
2( n − 1) ⋅ a ⎣ ( x + a )
(x + a ) ⎦
x
2n − 3
dx
=
+
2
2
2 n −1
2 ∫
2
2( n − 1) ⋅ a ⋅ ( x + a )
2( n − 1) ⋅ a ( x + a 2 ) n −1
2
dx
1
x−a
=
⋅ ln
+C
2
−a
2a
x+a
dx
1
1
1
证明：∫ 2
=
[
−
] dx
2
∫
2a x − a x + a
x −a
1
1
1
1
=
dx −
dx
∫
∫
2a x − a
2a x + a
1
1
=
⋅ ln x − a −
⋅ ln x + a + C
2a
2a
1
x−a
=
⋅ ln
+C
2a
x+a
∫x
2
2
（四）含有 ax + b (a > 0) 的积分（22~28）
⎧ 1
⋅ arctan
⎪
ab
dx
⎪
22. ∫ 2
=⎨
ax + b ⎪ 1
⋅ ln
⎪ 2 − ab
⎩
证明：
a
⋅x+C
b
(b > 0)
a ⋅x− −b
+C
a ⋅x+ −b
1
1.当 b > 0 时 ， 2
=
ax + b
(b < 0 )
1
1
1
=
⋅
b a
a
b
x2 +
x2 + ( )2
a
a
dx
1
1
∴ ∫ 2
= ∫
dx
ax + b a 2
b 2
x +( )
a
=
1
(a > 0 )
⋅
1 a
a
⋅
⋅ arctan
⋅x+C
a b
b
a
⋅x+C
b
ab
1
1
1
1
1
2.当 b < 0 时 ， 2
=
⋅ =
⋅
b a
a
ax + b
b
x 2 − (− )
x2 − ( − )2
a
a
=
∴
∫ ax
1
⋅ arctan
dx
1
= ∫
2
+b a
1
b
x − ( − )2
a
dx
2
−b
x−
1
a
=
⋅ ⋅ ln
+C
−b a
−b
2
x+
a
a
1
=
1
2 − ab
⋅ ln
a ⋅ x − −b
+C
a ⋅x+ −b
⎧ 1
⋅ arctan
⎪
dx
⎪ ab
综合讨论1 , 2 得：∫ 2
=⎨
ax + b ⎪ 1
⋅ ln
⎪ 2 − ab
⎩
23.
a
⋅x+C
b
(b > 0 )
a ⋅x− −b
+C
a ⋅x+ −b
(b < 0 )
x
1
dx =
⋅ ln ax 2 + b + C
( a > 0)
2a
+b
x
1
1
证明：∫ 2
dx = ∫ 2
dx 2
2 ax + b
ax + b
1
1
=
d (ax 2 + b )
2
∫
2 a ax + b
1
=
⋅ ln ax 2 + b + C
2a
∫ ax
2
- 11 -
x2
x b
dx
24. ∫ 2
dx = − ∫ 2
( a > 0)
ax + b
a a ax + b
x2
b
ax 2
1
证明：∫ 2
dx = ∫ 2
⋅ dx
a ax + b b
ax + b
b 1
1
= ∫( − 2
)dx
a b ax + b
b 1
b
1
= ∫ dx − ∫ 2
dx
a b
a ax + b
x b
dx
= − ∫ 2
a a ax + b
25 .
∫
dx
1
x2
=
⋅
ln
+C
x ( ax 2 + b ) 2b
ax 2 + b
x
dx
( ax 2 + b )
1
1
= ∫ 2
dx 2
2
2 x ( ax + b )
1
A
B
设： 2
= 2 +
2
2
x ( ax + b ) x
ax + b
证明：∫
dx
=
x ( ax 2 + b )
(a > 0)
∫x
2
则 1 = A( ax 2 + b ) + Bx 2 = x 2 ( Aa + B ) + Ab
1
⎧
A
=
⎪⎪
⎧ Aa + B = 0
b
∴ 有 ⎨
⇒
⎨
⎩ Ab = 1
⎪B = − a
⎪⎩
b
dx
1
1
a
于是 ∫
= ∫[ 2 −
]dx 2
2
2
x ( ax + b ) 2 bx
b ( ax + b )
1
1
a
1
=
dx 2 −
dx 2
∫
2
∫
2
2b x
2b ax + b
1
1
1
1
=
dx 2 −
d ( ax 2 + b )
∫
2
∫
2
2b x
2b ax + b
1
1
=
·ln x 2 −
· ln ax 2 + b + C
2b
2b
2
1
x
=
·ln
+C
2b
ax 2 + b
- 12 -
26 .
dx
1
a
dx
=−
− ∫ 2
( a > 0)
2
bx b ax + b
( ax + b )
1
A
B
证明：设： 2
= 2 +
2
2
x ( ax + b ) x
ax + b
∫x
2
则 1 = A( ax 2 + b ) + Bx 2 = x 2 ( Aa + B ) + Ab
1
⎧
A
=
⎪⎪
⎧ Aa + B = 0
b
∴ 有 ⎨
⇒
⎨
⎩ Ab = 1
⎪B = − a
⎪⎩
b
dx
1
a
于是 ∫ 2
= ∫[ 2 −
]dx
2
x ( ax + b )
bx
b ( ax 2 + b )
1 1
a
1
= ∫ 2 dx − ∫ 2
dx
b x
b ax + b
1
a
dx
=−
− ∫ 2
bx b ax + b
ax 2 + b
dx
a
1
27 . ∫ 3
=
ln
−
+C
2
2
2
x ( ax + b ) 2b
x
2bx 2
dx
x
证明：∫ 3
=
dx
x ( ax 2 + b ) ∫ x 4 ( ax 2 + b )
1
1
= ∫ 4
dx 2
2
2 x ( ax + b )
1
A
B
C
设： 4
=
+
+
x ( ax 2 + b ) x 2 x 4 ax 2 + b
(a > 0)
则 1 = Ax 2 ( ax 2 + b ) + B ( ax 2 + b ) + Cx 4
= ( Aa + C ) x 4 + ( Ab + Ba ) x 2 + Bb
⎧
1
⎪B = b
⎧ Aa + C = 0
⎪
a
⎪
⎪
∴ 有 ⎨ Ab + Ba = 0
⇒
⎨A = − 2
b
⎪ Bb = 1
⎪
⎩
2
⎪
a
⎪C = 2
b
⎩
dx
a
1
1
1
a2
1
2
2
于是 ∫ 3
= − 2 ∫ 2 dx +
dx +
dx 2
2
4
2 ∫
2
∫
2b x
x ( ax + b )
2b
x
2b ax + b
a
1
a
= − 2 ·ln x 2 −
+
· ln ax 2 + b + C
2
2
2b
2bx
2b
2
ax + b
a
1
=
ln
−
+C
2
2
2b
x
2bx 2
- 13 -
dx
x
1
dx
=
+
(a > 0 )
2 b ∫ ax 2 + b
+ b)2
2 b ( ax 2 + b )
dx
1
1
1
1
1
1
证明：∫
= −∫
d
= −
⋅
+ ∫
d
2
2
2
2
2
( ax + b )
2 ax
ax + b
2 ax ax + b
ax + b
2 ax
1
1
1
1
= −
⋅
− ∫
⋅
dx
2
2
2 ax ax + b
ax + b 2 ax 2
1
A
B
设：
=
+
，则 1 = A ( ax 2 + b ) + 2 Bax 2 = ( Aa + 2 Ba ) x 2 + Ab
2
2
2
2
2 ax ( ax + b )
2 ax
ax + b
1
⎧
A =
⎪
⎧ Aa + 2 Ba = 0
⎪
b
∴ 有 ⎨
⇒
⎨
⎩ Ab = 1
⎪B = − 1
2b
⎩⎪
1
1
1
于是 上式 = −
− ∫(
−
) dx
2
2
2 ax ( ax + b )
2 abx
2 b ( ax 2 + b )
1
1
1
1
1
= −
−
dx +
dx
2
∫
2
∫
2 ax ( ax + b ) 2 ab x
2 b 2 b ( ax 2 + b )
28 .
∫ ( ax
2
1
1
1
+
+
2
2 ax ( ax + b )
2 abx
2b
x
1
dx
=
+
2 b ∫ ax 2 + b
2 b ( ax 2 + b )
= −
1
∫ 2 b ( ax
2
+ b)
dx =
ax 2 + b − b
1
+
2
2 abx ( ax + b )
2b
1
∫ 2 b ( ax
2
+ b)
dx
（五）含有 ax 2 + bx + c (a > 0) 的积分（29~30）
2
2ax + b
⎧
⋅ arctan
+C
⎪
2
2
4
ac
−
b
4ac
−
b
dx
⎪
29. ∫ 2
=⎨
ax + bx + c ⎪
1
2ax + b − b 2 − 4 ac
⋅ ln
+C
2
⎪ b 2 − 4ac
2ax
+
b
+
b
−
4
ac
⎩
1
证明：∵ ax 2 + bx + c =
( 2ax + b ) 2 + ( 4ac − b 2 )
4a
dx
1
∴∫ 2
= 4a ∫
dx
2
ax + bx + c
( 2ax + b) + (4ac − b 2 )
[
(b 2 < 4ac)
( a > 0)
2
(b > 4ac)
]
dx
1
= 4a ∫
dx
2
ax + bx + c
( 2ax + b) + ( 4ac − b 2 ) 2
4a
1
dx
1
x
=
d (2ax + b)
公式 19 ：∫ 2
= ⋅ arctan + C
∫
2
2
x +a
a
a
2a ( 2ax + b ) + ( 4ac − b 2 ) 2
1.当 b 2 < 4ac 时 ，
∫
2
=
dx
1
=
⋅ ln
2a
x2 − a2
2ax + b
4ac − b 2
+C
2
=
- 14 -
4ac − b 2
⋅ arctan
dx
1
= 4a ∫
dx
2
ax + bx + c
( 2ax + b) + (4ac − b 2 )
1
= 4a ∫
dx
2
( 2ax + b) − (b 2 − 4ac)
4a
1
x−a
=
d (2ax + b )
∫
+C
2
2a ( 2ax + b) − ( b 2 − 4 ac ) 2
x+a
2.当 b 2 > 4ac 时 ，∫
公式 21：∫
2
1
b 2 − 4ac
⋅ ln
2ax + b − b 2 − 4ac
2ax + b + b 2 − 4 ac
+C
2
2ax + b
⎧
⋅ arctan
+C
⎪
2
2
4
ac
−
b
4ac
−
b
dx
⎪
综合讨论 1 , 2 得： ∫ 2
=⎨
ax + bx + c ⎪
1
2ax + b − b 2 − 4ac
⋅ ln
+C
2
⎪ b 2 − 4ac
2ax
+
b
+
b
−
4
ac
⎩
(b 2 < 4ac)
(b 2 > 4ac )
30.
x
1
b
dx
dx =
⋅ ln ax 2 + bx + c −
( a > 0)
2
∫
+ bx + c
2a
2a ax + bx + c
x
1 2ax + b − b
证明：∫ 2
dx = ∫
⋅
dx
2 a ax 2 + bx + c
ax + bx + c
1
2ax + b
1
−b
=
dx +
dx
∫
2
∫
2
2a ax + bx + c
2 a ax + bx + c
1
1
b
1
=
d (ax 2 + bx + c ) −
dx
2
2
∫
∫
2a ax + bx + c
2a ax + bx + c
1
b
dx
=
⋅ ln ax 2 + bx + c −
∫
2
2a
2 a ax + bx + c
∫ ax
2
（六）含有
31.
∫
dx
2
x +a
2
x 2 + a 2 ( a > 0 ) 的积分（31~44）
= arsh
x
+ C1 = ln ( x + x 2 + a 2 ) + C
a
证明：被积函数 f ( x ) =
( a > 0)
1
的定义域为{ x | x ∈ R}
x + a2
π
π
可令x = a tant ( − < t < ) , 则dx = d ( a tant ) = a sec 2 tdt , x 2 + a 2 =| a sect |
2
2
π
π
1
∵ − < t < , sect =
> 0 , ∴ x 2 + a 2 = a sect
2
2
cost
dx
1
∴ ∫
=∫
⋅ a sec 2 t dt
公式 87：∫ sectdt = ln | sect + tant | +C
2
2
a
sect
x +a
2
= ∫ sect dt
= ln sect + tant + C 2
在RtΔ ABC中，设 ∠B = t ,| BC |= a ,则 | AC | = x , | AB | =
∴ sect =
∴
∫
1
=
cost
dx
x2 + a2
x2 + a2
x
, tant =
a
a
= ln sect + tant + C 2
x2 + a2
= ln
x2 + a2 + x
+ C2
a
= ln
x 2 + a 2 + x − lna + C 2
= ln
x 2 + a 2 + x + C3
∵
x2 + a2 + x > 0
dx
∴ ∫
= ln ( x + x 2 + a 2 ) + C
2
2
x +a
- 15 -
32.
∫
dx
2
x
=
2 3
(x + a )
a
2
+C
x2 + a2
(a > 0)
1
证明：被积函数 f ( x) =
的定义域为{ x | x ∈ R}
( x 2 + a 2 )3
π
π
< t < ), 则dx = d (a tant ) = a sec 2 tdt , ( x 2 + a 2 ) 3 =| a 3 sec 3 t |
2
2
π
π
1
∵ − < t < , sect =
> 0 , ∴ ( x 2 + a 2 ) 3 = a 3 sec 3t
2
2
cost
dx
1
1
1
∴ ∫
=∫ 3
⋅ a sec 2 t dt = 2 ∫
dt
3
a sec t
a sect
(x 2 + a 2 )3
可令x = a tant
(−
1
1
cos tdt = 2 sint + C
2 ∫
a
a
=
在RtΔ ABC 中，设∠B = t , | BC | = a , 则 | AC | = x , | AB | =
| AC |
x
∴ sint =
=
2
| AB |
x + a2
∴
33.
∫
dx
∫
( x 2 + a 2 )3
x
x2 + a2
=
1
x
⋅ sint + C =
+C
2
a
a 2 x2 + a 2
dx = x 2 + a 2 + C
( a > 0)
证明：令 x 2 + a 2 = t (t > 0) , 则x = t 2 − a 2
1
−
1
t
∴ dx = (t 2 − a 2 ) 2 ⋅ 2tdt =
dt
2
t 2 − a2
∴
∫
x
x2 + a2
dx = ∫
t2 − a2
t
⋅
dt
t
t 2 − a2
= ∫ dt = t + C
将t = x 2 + a 2 代入上式得：
∫
34.
∫
x
( x 2 + a 2 )3
证明：∫
dx = −
x
(x 2 + a 2 )3
1
x2 + a2
x
2
x +a
+C
2
dx = x 2 + a 2 + C
( a > 0)
−
3
dx = ∫ x ⋅ ( x 2 + a 2 ) 2 dx =
3
−
1
(x 2 + a 2 ) 2 d ( x 2 + a 2 )
∫
2
3
1−
1
1
= ×
⋅ (x 2 + a 2 ) 2 + C
3
2
1−
2
1
=−
+C
x2 + a2
=
- 16 -
3
−
1
( x 2 + a 2 ) 2 dx 2
∫
2
x2 + a 2
35.
∫
x2
dx =
x 2 + a2
x
a2
⋅ x2 + a2 −
ln ( x + x 2 + a 2 ) + C
2
2
x2
证明：∫
x2 + a2
x2 + a2 − a2
dx = ∫
x2 + a 2
dx
= ∫ x 2 + a 2 dx − a 2 ∫
∵
1
∫
∴
36.
∫
2
x +a
x2
x2
2 3
dx
(公式39)
(公式31)
d x = ln ( x + x 2 + a 2 ) + C
x
dx = −
2
(x + a )
x +a
+ ln ( x + x 2 + a 2 ) + C
2
x2
证明：被积函数 f ( x) =
( x 2 + a 2 )3
的定义域为{ x | x ∈ R}
π
π
1
∵ − < t < , sect =
> 0, ∴
2
2
cost
∫
( a > 0)
π
π
x2
a 2 tan 2 t
2
( − < t < ), 则dx = d (a tant ) = a sec tdt ，
= 3
3
2
2
( x 2 + a 2 ) 3 | a sec t |
可令x = a tant
∴
x + a2
x
a2
2
2
dx = ⋅ x + a +
ln ( x + x 2 + a 2 ) − a 2 ⋅ ln ( x + x 2 + a 2 ) + C
2
2
x2 + a 2
x
a2
= ⋅ x2 + a 2 −
⋅ ln ( x + x 2 + a 2 ) + C
2
2
∫
2
2
1
2
x
a2
⋅ x2 + a2 +
⋅ ln ( x + x 2 + a 2 ) + C
2
2
x 2 + a 2 dx =
∫
( a > 0)
x2
(x 2 + a 2 )3
dx = ∫
tan 2 t
=
3
( x 2 + a 2 ) 3 a sec t
x2
tan 2 t
tan 2 t
sec 2 t − 1
2
⋅
a
sec
tdt
=
dt
=
∫ sect
∫ sect dt
a sec 3 t
1
dt = ∫ sectdt − ∫ cos tdt
sect
公式 87：∫ sec t dt = ln | sect + tant | +C
= ln sect + tant − sint + C 1
= ∫ sect dt − ∫
在RtΔ ABC 中，设∠B = t , | BC | = a , 则 | AC | = x , | AB | =
x
∴ sint =
∴
∫
x2 + a2
x2
( x 2 + a 2 )3
∵
∴
, tant =
x
1
, sect =
=
a
cost
x2 + a 2
x2 + a2
a
dx = ln sect + tant − sint + C 1
= ln
x2 + a2 + x
−
a
= ln
x 2 + a2 + x −
x
x2 + a2
x
2
x + a2
+ C1
− lna + C 1
x2 + a2 + x > 0
∫
x2
2
2 3
(x + a )
dx = −
x
2
x +a
2
+ ln ( x + x 2 + a 2 ) + C
- 17 -
37.
∫ x⋅
dx
x2 + a2
=
1
x2 + a2 − a
⋅ ln
+C
a
x
证明：令 x 2 + a 2 = t
( a > 0)
(t > 0) , 则x = t 2 − a 2
1
−
1
t
∴ dx = (t 2 − a 2 ) 2 ⋅ 2tdt =
dt
2
2
t − a2
dx
1
t
∴∫
=∫
⋅
dt
2
2
2
2
2
x⋅ x + a
t⋅ t −a
t − a2
1
公式 21：∫
=∫ 2
dt
t − a2
1
t−a
=
⋅ ln
+C
2a
t+a
=
1
(t − a ) 2
⋅ ln 2
+C
2a
t − a2
将 t = x 2 + a 2 代入上式得：
∫
dx
x ⋅ x2 + a2
提示：
log a b n = nlog a b
38.
dx
dx
1
x−a
=
⋅ ln
+C
2a
x+a
x2 − a2
=
1
( x 2 + a 2 − a) 2
⋅ ln
+C
2a
x2 + a2 − a2
=
1
( x 2 + a 2 − a) 2
⋅ ln
+C
2a
x2
=
1
x2 + a2 − a
⋅ ln
+C
a
x
x2 + a2
+C
a2 x
( a > 0)
⋅ x2 + a2
dx
1
1
证明：∫
= −∫
d
x2 ⋅ x 2 + a 2
x2 + a2 x
1
1
令t =
(t ≠ 0) , 则x =
x
t
1
1
1
t
∴ −∫
d = −∫
dt = − ∫
dt
1
x2 + a2 x
1 + a2t2
2
+a
t2
1
2a 2 t
=− 2 ∫
dt
2a
1 + a 2t 2
∫x
2
=−
=−
1
2a 2
∫
1
2 2
d (1 + a 2 t 2 )
1+ a t
1
1−
1
1
=− 2 ⋅
(1 + a 2 t 2 ) 2 + C
1
2a
1−
2
1
= − 2 ⋅ 1+ a2t 2 + C
a
将t =
- 18 -
1
dx
x2 + a2
代入上式得：
=
−
+C
∫ x2 ⋅ x2 + a2
x
a2 x
39 .
∫
x 2 + a 2 dx =
证法 1： ∵ ∫
x
2
x2 + a2 +
x 2 + a 2 dx = x
∴
1
公式 31：
∫
x + a dx +
∫
又
x2+ a2
2
∫
x2+ a2 − ∫
39 .
∫
∫
2
x +a
x2
2
x2+ a2
x 2 + a 2 dx = x
x 2 + a 2 dx =
x
2
x2
x2+ a2
dx =
x2+ a2 +
( a > 0)
x2+ a2
dx
①
x2 + a2
dx = x
x 2 + a 2 ) + C (a > 0)
由① + ②得， 2 ∫
即
x2
x 2 + a 2 dx − ∫
dx = ln ( x +
x2 + a2 ) + C
x2+ a2 − ∫ x d
=x
2
a2
⋅ ln ( x +
2
∫
a2
x2 + a2
= a 2 ⋅ ln ( x +
x 2 + a 2 ) + C1
x 2 + a 2 + a 2 ⋅ ln ( x +
a2
⋅ ln ( x +
2
②
dx
x2 + a2 )
x2 + a2 ) + C
x
a2
2
2
⋅
x
+
a
+
⋅ ln ( x + x 2 + a 2 ) + C
(a > 0)
∫
2
2
π
π
证法 2： 令 x = a ⋅ tant ( − < t < ) ，则 x 2 + a 2 = a 1 + tan 2 t = a·sect ,
2
2
π
π
1
∵ − < t < , sec t =
> 0 , ∴ x 2 + a 2 = a·sect
2
2
cos t
提示：1 + tan 2 t = sec 2 t
x 2 + a 2 dx =
∴ ∫ x 2 + a 2 dx = ∫ a·sectd ( a·tant ) = a 2 ∫ sect dtant
= a 2 sect ⋅ tant − a 2 ∫ tantdsect
sin 2 t
∫ cos 3 t dt
1 − cos 2 t
1
1
1
=∫
dt = ∫
⋅
dt − ∫
dt
3
2
cos t
cost cos t
cost
= ∫ sect dtant − ∫ sectdt
①
又 ∫ tantdsect = ∫ tant ⋅ sect ⋅ tantdt =
1 2
( a sect ·tant + a 2 ∫ sectdt )
2
又 ∫ sectdt = ln | sect + tant | + C 1 （公式 87 ）
联立①②有 a 2 ∫ sect dtant =
1 2
1
a sect ⋅ tant + a 2 ln | sect + tant | + C 2
2
2
∵ x = a ⋅ tant ,∴ 在 Rt Δ ABC 中，可设 ∠ B = t ,| BC |= a ，
联立③④有 a 2 ∫ sect dtant =
则 | AC |= a·tant = x,| AB |=
∴ sect =
1
=
cost
②
③
④
⑤
a2+ x2
a2+ x2
x
, tant =
a
a
1 2
1
x
a2
x+ x2 + a 2
a sect ·tant + a 2 ln | sect + tant | = ⋅ x 2 + a 2 +
⋅ ln |
|
2
2
2
2
a
x
a2
a2
= ⋅ x2 + a2 +
⋅ ln ( x + x 2 + a 2 ) −
⋅ lna
2
2
2
x
a2
2
2
2
2
综合①②③④⑤得
x
+
a
dx
=
⋅
x
+
a
+
⋅ ln ( x + x 2 + a 2 ) + C
∫
2
2
∴
- 19 -
40.
∫
( x 2 + a 2 ) 3 dx =
x
3
⋅ ( 2 x 2 + 5a 2 ) x 2 + a 2 + ⋅ a 4 ⋅ ln ( x + x 2 + a 2 ) + C
8
8
(a > 0)
证明：被积函数 f ( x ) = ( x 2 + a 2 ) 3 的定义域为{x | x ∈ R}
π
π
< t < ), 则 ( x 2 + a 2 ) 3 =| a 3 sec 3 t |
2
2
π
π
1
∵ − < t < , sect =
> 0 , ∴ ( x 2 + a 2 ) 3 = a 3 ⋅ sec 3 t
2
2
cost
可令 x = a tant
∴
∫
(−
( x 2 + a 2 ) 3 dx = ∫ a 3 ⋅ sec 3 t d (a tant ) = a 4 ∫ sec 3 t d tant
= a 4 sec 3 t ⋅ tant − a 4 ∫ tant d sec 3 t
= a 4 sec 3 t ⋅ tant − a 4 ∫ tant ⋅ 3 ⋅ sec 2 t ⋅ sect tant dt
= a 4 sec 3 t ⋅ tant − 3a 4 ∫ tan 2 t ⋅ sec 3t dt
= a 4 sec 3 t ⋅ tant − 3a 4 ∫ tan 2 t ⋅ sect d tant
= a 4 sec 3 t ⋅ tant − 3a 4 ∫ ( sec 2 t − 1) ⋅ sect d tant
= a 4 sec 3 t ⋅ tant − 3a 4 ∫ sec 3 t d tant + 3a 4 ∫ sect d tant
1 4 3
( a sec t tant + 3a 4 ∫ sect d tant )
4
∵ ∫ sect d tant = sect ⋅ tant − ∫ tant d sect
移项并整理的：a 4 ∫ sec 3t d tant =
①
= sect ⋅ tant − ∫ tan 2 t ⋅ sectdt
= sect ⋅ tant − ∫ ( sec 2 t − 1) ⋅ sectdt
②
= sect ⋅ tant − ∫ sec 3 t dt + ∫ sect dt
又 ∵ ∫ sect d tant = ∫ sec 3 t dt
③
公式 87：∫ sec t dt = ln | sect + tant | +C
1
1
⋅ sect ⋅ tant + ∫ sect dt
2
2
1
1
= ⋅ sect ⋅ tant + ln sect + tant + C1
④
2
2
1
3
3
联立①④得 a 4 ∫ sec 3 t d tant = a 4 sec 3 t ⋅ tant + a 4 sect ⋅ tant + a 4 ⋅ ln sect + tant + C1
4
8
8
联立②③得：a 4 ∫ sec 3 t d tant =
在RtΔABC 中，设 ∠B = t , | BC | = a , 则 | AC | = x , | AB | =
x
1
∴ tant = , sect =
=
a
cos t
∴ a 4 ∫ sec 3 t d tant =
- 20 -
∫
x2 + a2
a
a4 x x2 + a2
3a 4
x 2 + a2 x 3 4
2
2
⋅ ⋅
⋅
x
+
a
+
⋅
⋅ + a ⋅ ln
4 a
8
a
a 8
a3
x2 + a 2 + x
+ C1
a
x 2
3a 2 ⋅ x
3a 4
(x + a 2 ) x2 + a 2 +
⋅ x2 + a 2 +
⋅ ln x 2 + a 2 + x + C
4
8
8
x
3
( x 2 + a 2 ) 3 dx = ⋅ ( 2 x 2 + 5a 2 ) x 2 + a 2 + ⋅ a 4 ⋅ ln ( x + x 2 + a 2 ) + C
8
8
=
∴
x2 + a2
41.
∫x⋅
x 2 + a 2 dx =
1
(x 2 + a 2 ) 3 + C
3
( a > 0)
1
1
证明：∫ x ⋅ x + a dx = ∫ ( x 2 + a 2 ) 2 dx 2
2
1
1
2
2 2
= ∫ (x + a ) d (x 2 + a 2 )
2
1
1+
1
1
= ×
⋅ (x 2 + a 2 ) 2 + C
1
2
1+
2
1
=
(x 2 + a 2 )3 + C
3
2
2
- 21 -
x
a4
2
2
2
2
42. ∫ x ⋅ x + a dx = ⋅ (2 x + a ) x + a −
⋅ ln ( x + x 2 + a 2 ) + C
8
8
2
2
2
(a > 0)
证明：被积函数 f ( x ) = x 2 ⋅ x 2 + a 2 的定义域为{ x | x ∈ R}
π
π
可令 x = a tant (− < t < ), 则 x 2 ⋅ ( x 2 + a 2 ) = a 2 tan 2 t | a sect |
2
2
π
π
1
∵ − < t < , sect =
> 0 , ∴ x 2 ⋅ ( x 2 + a 2 ) = a 3 tan 2 t ⋅ sect
2
2
cost
∫x
∴
2
⋅ ( x 2 + a 2 ) dx = ∫ a 3tan 2 t ⋅ sect d ( a tant ) = a 4 ∫ tan 2 t ⋅ sect d tant = a 4 ∫ tan 2 t ⋅ sec 3 t dt
= a 4 ∫ tant ⋅ sec 2 t dsect
= a 4 ∫ tant ⋅ (1 + tan 2 t ) dsect = a 4 ∫ tant dsect + a 4 ∫ tan 3t dsect
= a 4 ∫ tant dsect + a 4 ⋅ tan 3t ⋅ sect − a 4 ∫ sect dtan 3 t
= a 4 ∫ tant dsect + a 4 ⋅ tan 3t ⋅ sect − 3a 4 ∫ sec 3t tan 2 t dt
= a 4 ∫ tant dsect + a 4 ⋅ tan 3t ⋅ sect − 3a 4 ∫ sec 2 t tant dsect
1 4
( a ∫ tant dsect + a 4 ⋅ tan 3 t ⋅ sect )
4
a4
a4
=
tant dsect +
⋅ tan 3t ⋅ sect
∫
4
4
∵ ∫ tant d sect = sect ⋅ tant − ∫ sect dtant = sect ⋅ tant − ∫ sec 3 tdt
移项并整理的：a 4 ∫ tant ⋅ sec 2 t dsect =
①
= sect ⋅ tant − ∫ (1 + tan 2 t ) ⋅ sectdt
= sect ⋅ tant − ∫ sect dt − ∫ tan 2 t sect dt
= sect ⋅ tant − ∫ sect dt − ∫ tant d sect
1
1
⋅ sect ⋅ tant − ∫ sect dt 公式 87：∫ sect dt = ln | sect + tant | +C
2
2
1
1
= ⋅ sect ⋅ tant − ln sect + tant + C1
②
2
2
a4
a4
a4
4
2
联立①②得： a ∫ tant ⋅ sec t dsect =
sect ⋅ tant −
ln sect + tant +
tan 3 t ⋅ sect + C1
8
8
4
移项并整理得：∫ tant d sect =
在RtΔABC 中，设 ∠B = t , | BC | = a , 则 | AC | = x , | AB | =
∴ tant =
x
1
, sect =
=
a
cos t
x2 + a2
a
a 4 x x2 + a 2 a 4
∴ a ∫ tant ⋅ sec t dsect =
⋅ ⋅
−
⋅ ln
8 a
a
8
4
2
x2 + a2
x2 + a 2 + x
a 4 x3
x2 + a2
+
⋅ ⋅
+ C1
a
4 a3
a
a4x
a4
x3
2
2
2
2
=
⋅ x +a −
⋅ ln x + a + x +
⋅ x2 + a 2 + C2
8
8
4
4
x
a
= ⋅ (2 x 2 + a 2 ) x 2 + a 2 −
⋅ ln x 2 + a 2 + x + C
8
8
4
a
a4
∵ x 2 + a 2 + x > 0 ，∴
⋅ ln x 2 + a 2 + x =
⋅ ln ( x + x 2 + a 2 )
8
8
4
x
a
∴ ∫ x 2 ⋅ x 2 + a 2 dx = ⋅ (2 x 2 + a 2 ) x 2 + a 2 −
⋅ ln ( x + x 2 + a 2 ) + C
8
8
- 22 -
43 .
∫
x2 + a2
dx =
x
x2 + a2 − a
+C
x
x 2 + a 2 + a ⋅ ln
x2 + a2
的定义域为 { x | x ≠ 0}
x
证明： 被积函数 f ( x ) =
令 x2 + a2 = t
(t ≥ 0 且 t ≠ a ) , 则 x =
1
∴ dx =
∴
∫
t
∫
2
t − a2
⋅
t
t2 − a2
dt
dt =
∫t
2
t2
dt
− a2
t2 − a2
t2 − a2
t2 − a2 + a2
1
公式 21 ：∫
=∫
dt = ∫ dt + a 2 ∫ 2
dt
2
2
2
t −a
t −a
1
t−a
a
(t − a ) 2
= t + a2 ⋅
⋅ ln
+ C = t + ⋅ ln 2
+C
2a
t+a
2
t − a2
将t =
44 .
t
−
1 2
( t − a 2 ) 2 ⋅ 2 tdt =
2
x2 + a2
dx =
x
∫
(a > 0)
x2 + a2
dx =
x
x 2 + a 2 代入上式得： ∫
x2 + a2
dx = −
x2
x2 + a2
+ ln ( x +
x
x2 + a2 +
a
( x 2 + a 2 − a) 2
⋅ ln
+C
2
x2 + a2 − a2
=
x 2 + a 2 + a ⋅ ln
=
x 2 + a 2 + a ⋅ ln
x2 + a2 ) + C
dx
1
x −a
=
⋅ ln
+ C
2a
x +a
x2 − a2
( x 2 + a 2 − a)
+C
x
x2 + a2 − a
+C
x
(a > 0)
x2 + a2
的定义域为 { x | x ≠ 0}
x2
π
1. 当 x > 0时 , 可令 x = a tant ( 0 < t < ) , 则 dx = d ( a tant ) = a sec 2 tdt ,
2
a sect
x2 + a2
π
1
x2 + a2
sect
=
，
∵
0
<
t
<
,
sect
=
>
0
,
∴
=
2
2
2
2
2
cost
x
a tan t
x
a tan 2 t
证明：被积函数
∴
∫
f ( x) =
x2 + a2
dx =
x2
sect
∫ a tan
2
t
⋅ a sec 2 tdt =
sect
∫ tan
2
t
⋅ (1 + tan 2 t ) dt
sect
1
cos 2 t
⋅
dt
=
sect
dt
+
⋅
∫ tan 2 t
∫
∫ cost sin 2 t dt
cost
1
= ∫ sect dt + ∫
dt = ∫ sect dt + ∫
dsint
2
sin t
sin 2 t
公式 87：∫ sec t dt = ln | sect + tant | +C
1
= ln sect + tant −
+ C1
sint
=
∫ sect dt +
在 Rt Δ ABC 中，设 ∠ B = t , | BC | = a , 则 | AC | = x , | AB | =
∴ sint =
∴
∫
x
x2 + a2
x2 + a2
dx = ln
x2
= −
∵
, tant =
x2 + a2 + x
−
a
x2 + a2
+ ln
x
x2 + a2 + x > 0
2. 当 x < 0时 , 同理可证得：
综合讨论 1 , 2 得： ∫
x
1
, sect =
=
a
cost
x2 + a2
a
x2 + a2
+ C1
x
x 2 + a 2 + x − lna + C 1
x2 + a2
dx = −
x2
∴
∫
∫
x2 + a2
dx = −
x2
x2 + a2
dx = −
x2
x2 + a2
x2 + a2
+ ln ( x 2 + a 2 + x ) + C
x
x2 + a2
+ ln ( x 2 + a 2 + x ) + C
x
x2 + a2
+ ln ( x 2 + a 2 + x ) + C
x
- 23 -
（七）含有 x 2 − a 2 (a > 0) 的积分（45~58）
45.
∫
dx
x2 − a2
=
x
| x|
⋅ arsh
+ C1 = ln | x + x 2 − a 2 | + C
|x|
a
1
证法1：被积函数 f(x) =
x2 − a2
( a > 0)
的定义域为{x | x > a或 x < − a}
π
1 . 当 x > a 时 ,可设 x = a⋅ sect ( 0 < t < ) ，则 dx = a⋅ sect⋅ tantdt
2
π
x 2 − a 2 = a sec 2 t − 1 = a⋅ tant ∵ 0 < t < ， x 2 − a 2 = a⋅ tant
2
dx
a⋅ sect⋅ tant
∴∫
=∫
dt = ∫ sectdt 公式 87：∫ sec tdt = ln | sec t + tan t | +C
a⋅ tant
x2 − a 2
= ln | sect + tant | + C 2
在 Rt ΔABC 中，可设 ∠B = t ,| BC |= a ，
则 | AB |= x ,| AC |=
∴ sec t =
∴∫
x2 − a 2
x2− a2
a
1
x
| AC |
= , tan t =
=
cos t a
| BC |
x+ x 2 − a 2
= ln | sec t + tan t |= ln |
|
a
x2 − a2
dx
= ln | x + x 2 − a 2 | +C 3
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
由讨论 1可知
∫
dx
2
x −a
2
= −∫
dμ
2
μ −a
2
= − ln | μ + μ 2 − a 2 | + C 4
= − ln | − x + x 2 − a 2 | + C 4 = ln
1
| − x+ x 2 − a 2 |
+ C4
| − x+ x 2 − a 2 |
= ln
+ C4
a2
= ln | − x − x 2 − a 2 | + C 5
综合讨论 1 , 2 ，可写成
- 24 -
∫
dx
x2− a2
=
x
|x|
⋅ arsh
+ C1 = ln | x + x 2 − a 2 | + C
|x|
a
45.
∫
dx
2
x −a
2
=
x
|x|
⋅ arsh
+ C1 = ln | x+ x 2 − a 2 | + C
| x|
a
1
证法2：被积函数 f(x) =
2
x − a2
( a > 0)
的定义域为 {x | x > a或x < −a}
1 . 当 x > a 时 ,可设 x = a⋅ cht (t > 0)，则 t = arch
x
a
x 2 − a 2 = a 2 ch2 t − a 2 = a⋅ sht , dx = a⋅ shtdt
dx
a⋅ sht
∴∫
=∫
dt = ∫ dt = t + C1
a⋅ sht
x2 − a 2
2
⎡x
⎤
x
x⎞
⎛
= arch + C = ln ⎢ + ⎜ ⎟ − 1 ⎥ + C2
a
⎢a
⎥
⎝a⎠
⎣
⎦
= ln | x + x 2 − a 2 | + C3
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
由讨论 1可知
∫
dx
2
x −a
2
= −∫
dμ
2
μ −a
2
= − ln | μ+ μ 2 − a 2 | + C4
= − ln( − x+ x 2 − a 2 ) + C4 = ln
= ln
1
| − x+ x 2 − a 2 |
+ C4
| − x+ x 2 − a 2 |
+ C4
a2
= ln | − x − x 2 − a 2 | + C5
综合讨论 1 , 2 ，可写成
∫
dx
x2 − a2
=
x
|x|
⋅ arsh
+ C1 = ln | x+ x 2 − a 2 | + C
| x|
a
- 25 -
46.
∫
dx
=−
(x 2 − a 2 )3
x
a 2 ⋅ x2 − a2
证明：被积函数 f(x) =
+C
1
( x 2 − a 2 )3
( a > 0)
的定义域为{ x | x > a或x < −a}
π
(0 < t < ) ，则 dx = a⋅ sect⋅ tantdt
2
π
( x 2 − a 2 ) 3 = a 3 ⋅ tan 3 t
∵ 0 < t < ，tant > 0 , ( x 2 − a 2 ) 3 = a 3 ⋅ tan 3 t
2
dx
a⋅ sect ⋅ tant
1 sect
∴ ∫
=∫ 3
dt = 2 ∫
dt
3
a ⋅ tan t
a tan 3 t
( x 2 − a 2 )3
1 . 当 x > a 时 ,可设 x = a⋅ sect
1 cos 2 t
1
∫ cos t ⋅ sin 2 t dt = a 2
1
∫ sin 2 t d sint
1
=− 2
+C
a sin t
1
a2
1
= 2
a
=
cos t
dt
2
t
∫ sin
在 Rt ΔABC 中，可设 ∠B = t ,| BC |= a ，
则 | AB |= x ,| AC |=
x2 − a2
x
∴ sin t =
∴
∫
dx
=−
(x2 − a2 )3
x
+C
a2 ⋅ x2 − a2
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
∴
∫
dx
( x 2 − a 2 )3
dμ
= −∫
由讨论 1可知 − ∫
(μ 2 − a2 )3
dμ
2
(μ − a )
47.
∫
x
2
x −a
证明：∫
2
dx =
x
x2 − a2
x2 − a2 + C
3
=
dx
将 μ = − x 代入得：∫
综合讨论 1 , 2 得： ∫
2
dx
2
(x − a )
3
=−
−
1
2
2
2 dx 2
(
x
−
a
)
∫
2
1
−
1
( x 2 − a 2 ) 2 d (x 2 − a 2 )
∫
2
1
1−
1
1
2
2
= ×
(x − a ) 2 + C
1
2
1−
2
= x2 − a2 + C
- 26 -
a2 ⋅ x2 − a2
a ⋅ x2 − a2
( a > 0)
=
x
x
1
dx =
a ⋅ (μ2 − a2 )
=−
( x 2 − a 2 )3
2
μ
2
2
+C
+C
+C
x2 − a2
48.
∫
x
2
2
(x − a )
1
dx = −
3
+C
2
x − a2
( a > 0)
x
证明：被积函数 f(x) =
的定义域为{ x | x > a或x < − a}
( x 2 − a 2 )3
π
1 . 当 x > a 时 ,可设 x = a⋅ sect ( 0 < t < ) ，则 dx = a⋅ sect⋅ tantdt
2
x
a⋅ sect
π
x
sect
= 3
∵0 < t < ，
= 2
3
2
2 3
2
2
3
2
a ⋅ tan 3 t
a ⋅ tan t
(x − a )
(x − a )
∴
∫
x
2
2
(x − a )
dx = ∫
3
sect
⋅ a⋅ sect ⋅ tant dt
a ⋅ tan 3 t
2
1 sec 2 t
1
1
dt = ∫
dt
2
∫
a tan t
a sin 2 t
1
1
= − ∫ − csc 2 tdt = − ⋅ cot t + C
a
a
=
在 Rt ΔABC 中，可设 ∠B = t ,| BC |= a ，
则 | AB |= x ,| AC |=
a
∴ cot t =
∴
x2 − a2
x2 − a2
x
1
a
1
dx = − ⋅
+C =−
+C
2
2
2
a x −a
(x − a )
x − a2
∫
2
2 3
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
∴
∫
x
( x 2 − a 2 )3
由讨论 1可知
2
2
(μ − a )
3
dμ = −
x
( x 2 − a 2 )3
x
综合讨论 1 , 2 得：∫
dμ
1
2
μ − a2
+C
1
dx = −
x2 − a2
+C
1
+C
(x − a )
x − a2
x
x 2 2 a
dx =
x − a + ⋅ ln x + x 2 − a 2 + C
( a > 0)
2
2
2
2
x −a
2
∫
( μ 2 − a 2 )3
μ
∫
将 μ = − x 代入得：∫
49.
μ
dx = ∫
证明：∫
2
2 3
dx = −
2
2
x2
x2 − a 2
dx = ∫
x2 − a2 + a2
x2 − a2
= ∫ ( x 2 − a2 +
dx
a2
x2 − a2
= ∫ x 2 − a 2 dx + a 2 ∫
)dx
1
x 2 − a2
dx
x
a2
2
2
∵ ∫ x − a dx = ⋅ x − a −
⋅ ln x + x 2 − a 2 + C
① (公式53)
2
2
dx
a2 ∫
= a 2 ⋅ ln x+ x 2 − a 2 + C
② (公式45)
2
2
x −a
x2
x 2 2 a2
∴ 由① + ②得：
dx
=
x − a + ⋅ ln x + x 2 − a 2 + C
∫ x 2 − a2
2
2
2
2
- 27 -
50 .
x2
∫
2
2
(x − a )
3
x
dx = −
2
x −a
+ ln x +
2
x2
证明 ：被积函数 f(x) =
x2 − a2 + C
( a > 0)
的定义域为 { x | x > a或 x < − a}
( x 2 − a 2 )3
π
) ，则 dx = a ⋅ sect ⋅ tantdt
2
x2
a 2 ⋅ sec 2 t
π
x2
sec 2 t
= 3
∵
0
<
t
<
，
∴
=
2
a ⋅ tan 3 t
a ⋅ tan 3 t
( x 2 − a 2 )3
(x2 − a2 )3
1 . 当 x > a 时 , 可设 x = a ⋅ sect
∴
x2
∫
(x2 − a2 )3
(0 < t <
sec 2 t
sec 3 t
1
cos 2 t
⋅
a
⋅
sect
⋅
tant
dt
=
dt
=
⋅
∫ tan 2 t
∫ cos 3 t sin 2 t dt =
a ⋅ tan 3 t
dx = ∫
∫ sin
2
1
dt
t ⋅ cos t
cos t
1
1
1
dt = ∫
d sin t = ∫ ( 2 +
) d sin t
2
2
2
t ⋅ cos t
sin t (1 − sin t )
sin t 1 − sin 2 t
1
1
1
1
=∫
d sin t + ∫
d sin t = ∫
d sin t − ∫
d sin t
2
2
2
2
sin t
1 − sin t
sin t
sin t − 1
1
1
1
1
=∫
d sin t + ∫ (
−
) d sin t
2
2 sin t + 1 sin t − 1
sin t
1
1
1
1
1
=∫
d sin t + ∫
d ( sin t + 1) − ∫
d ( sin t − 1)
2
2 sin t + 1
2 sin t − 1
sin t
1
1
1
=−
+ ln sin t + 1 − ln sin t − 1 + C1
sin t 2
2
1
1
sin t + 1
1
1
( sin t + 1）2
=−
+ ln
+ C1 = −
+ ln
+ C1
sin t 2
sin t − 1
sin t 2
sin 2 t − 1
=
∫ sin
=−
2
1
1
( sin t + 1）2
1
+ ln
⋅ ( −1) + C1 = −
+ ln tan t + sec t + C 2
2
sin t 2
sin t
cos t
x2− a2
在 Rt Δ ABC 中，可设 ∠ B = t ,| BC |= a ，
则 | AB |= x ,| AC |=
x2 − a2
, tan t =
x
∴ sin t =
∴
x2
∫
2
2 3
x
dx = −
2
(x − a )
x −a
2
+ ln
x2 − a2
x
, sec t =
a
a
x+ x 2 − a 2
+ C2 = −
a
x
2
x −a
2
+ ln x +
x2 − a2 + C
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
∴
∫
x2
(x2 − a2 )3
2
=−
n
提示：
loga b = n loga b
2
2
x
x2 − a2
x
2
x −a
2 3
2
dx = −
− ln
2 3
(x − a )
2
− ln μ +
x
dx = −
2
x −a
x
x2 − a2
− ln
x2 − a 2 − x2
2
μ2 − a2 + C
− ln − x +
x2 − a2 + C
( x 2 − a 2 − x )( x 2 − a 2 + x )
+C =−
( x 2 − a 2 + x)
+ ln x +
x2
2
μ −a
(x − a )
(x 2 − a 2 )3
综合讨论 1 , 2 得： ∫
μ
dμ =
x2
x2
=−
- 28 -
2 3
dμ
(μ − a )
将 μ = − x 代入得：∫
∫
( μ 2 − a 2 )3
μ2
由讨论 1可知 − ∫
=
μ2
dx = − ∫
x2 − a2 + x
x
x2 − a2
+ 2 ln a − ln
x2− a2 + C
dx = −
x
2
x −a
2
+ ln x +
x2 − a 2 + C
+C
1
x2 − a2 + x
+C
51.
1
a
= ⋅ arccos
+C
(a > 0)
2
2
a
|
x
|
x −a
1
证法 1：
被积函数 f(x) =
的定义域为 {x | x > a或x < −a}
2
2
x x −a
π
1. 当x > a时 ,可设x = a⋅ sect ( 0 < t < )，则
2
∫x
dx
x x 2 − a 2 = a 2 ⋅ sect sec2t − 1 = a 2 sect⋅ tant ,dx = a⋅ sect⋅ tant d t
dx
a ⋅ sect ⋅ tant
1
∴ ∫
=∫ 2
dt = ∫ dt
a sect⋅ tant
a
x x2 − a2
1
= t + C1
a
a
a
∵ x = a⋅ sect,∴ cost = , ∴ t = arccos
x
x
dx
1
a
∴ ∫
= ⋅ arccos + C
x
x x 2 − a2 a
2 .当x < − a ,即− x > a时，令 μ = − x ,即x = − μ
dx
dμ
1
a
由讨论 1可知 ∫
=∫
= ⋅ arccos + C2
μ
x x2 − a 2
μ μ2 − a2 a
1
a
= ⋅ arccos
+C
a
−x
dx
1
a
综合讨论1 , 2，可写成 ∫
= ⋅ arccos
+C
| x|
x x 2 − a2 a
- 29 -
51.
1
a
= ·arccos
+C
(a > 0)
| x|
x2 − a2 a
1
证法2：
被积函数 f(x) =
的定义域为 {x | x > a或x < −a}
x x2 − a 2
1. 当x > a 时 ,可设x = a⋅ ch t (0 < t ) ，则
∫x
dx
x x 2 − a 2 = a⋅ cht⋅ a⋅ sht = a 2 cht⋅ sht , dx = a⋅ sht dt
dx
a⋅ sht
1 1
∴ ∫
=∫
dt = ∫ ⋅
dt
a⋅ cht⋅ sht
a cht
x x2 − a2
1 cht
1
1
dt = ∫
dsht
2
∫
a ch t
a 1 + sh 2t
1
dx
1
x
= ⋅ arctan( sht ) + C
公式 19 ：∫ 2
= arctan + C
2
a
a
a
x +a
=
x
x2 − a2
2
∵ x = a⋅ cht, ∴ cht = , ∴ sht = 1 − ch t =
a
a
x2 − a2
,∠ B = y ,| BC |= a
a
在RtΔABC中，设 tany = sht =
∴ y = arctan (sht ), | AC |= x 2 − a 2, | AB |= | AC |2 + | BC |2 = x
∴ cosy =
| BC | a
=
| AB | x
即cosy = cos arctan (sht ) =
a
x
a
∴ arctan(sht ) = arccos + C
x
dx
1
1
a
∴ ∫
= ⋅ arctan (sht ) + C = ⋅ arccos + C
a
x
x x2 − a2 a
2 .当x < − a ,即 − x > a时，令 μ = − x ,即 x = − μ
dx
dμ
1
a
由讨论 1可知 ∫
=∫
= ⋅ arccos + C2
μ
x x2 − a2
μ μ 2 − a2 a
1
a
⋅ arccos
+C
a
−x
dx
1
a
综合讨论1 , 2 ，可写成 ∫
= ⋅ arccos
+C
|x|
x x2 − a2 a
=
- 30 -
52.
∫x
dx
2
x2 − a 2
=
x2 − a2
+C
a2 x
(a > 0)
1
证明：被积函数 f(x) =
的定义域为 {x | x > a或x < − a}
x2 − a2
1
1
1
1
t3
1 . 当 x > a 时 , 可设 x =
(0 < t < ) ，则 dx = − 2 dt ,
=
t
a
t
x 2 x2 − a 2
1 − a 2t 2
dx
t3
1
∴ ∫
=∫
⋅ ( − 2 ) dt
t
x 2 x2 − a2
1 − a 2t 2
x
2
t
= −∫
1 − a 2t 2
dt = −
1
−
1
2 2
2 dt 2
(
1
−
a
t
)
∫
2
1
1
1−
−
1
1
1
2 2
= 2 ∫ (1 − a t ) 2 d (1 − a 2 t 2 ) = 2 ⋅
⋅ (1 − a 2 t 2 ) 2 + C
1
2a
2a
1−
2
=
1 − a 2t 2
+C
a2
dx
1
1
将x = , 即 t = 代入上式得：∫
t
x
x2
x2 − a 2
1
1
1
⋅ 1− a2 ( )2 + C = 2 ⋅
2
x
a
a
=
∵ x>a>0 ∴
∫x
dx
x2 − a2
+C
x
1
⋅
a2
=
x2 − a2
+C
x2
x2 − a2
+C
a2 x
=
x2 − a2
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
由讨论 1可知 ∫
2
dx
x2 x2 − a2
将μ = − x代入上式得：∫
综合讨论 1 , 2 得：∫
= −∫
μ 2 μ2 − a2
dx
x2
x2 − a2
dx
x2
dμ
x2 − a2
=
=
=−
μ2 − a2
+C
a2 μ
x2 − a2
+C
a2 x
x2 − a2
+C
a2 x
- 31 -
53 .
∫
x 2 − a 2 dx =
x
a2
x2 − a2 −
⋅ ln x + x 2 − a 2 + C
2
2
( a > 0)
x 2 − a 2 的定义域为 { x | x > a 或 x < − a}
π
1 . 当 x > a 时 , 可设 x = a ⋅ sect ( 0 < t < ) ，则 x 2 − a 2 = a ⋅ tan t
2
π
∵ 0 < t < ，∴ x 2 − a 2 = a ⋅ tan t
2
证明 ：被积函数 f(x) =
∴
x 2 − a 2 dx = ∫ a ⋅ tan t d ( a ⋅ sect ) = a 2 ∫ tan t d sec t
∫
= a 2 ⋅ tan t ⋅ sec t − a 2 ∫ sec t d tan t
= a 2 ⋅ tan t ⋅ sec t − a 2 ∫ sec 3 t dt
= a 2 ⋅ tan t ⋅ sec t − a 2 ∫ sec t (1 + tan 2 t ) dt
= a 2 ⋅ tan t ⋅ sec t − a 2 ∫ sec t dt − a 2 ∫ sec t tan 2 t dt
= a 2 ⋅ tan t ⋅ sec t − a 2 ∫ sec t dt − a 2 ∫ tan t d sec t
= a 2 ⋅ tan t ⋅ sec t − a 2 ⋅ ln sec t + tan t − a 2 ∫ tan t d sec t
移项并整理得： a 2 ∫ tan t d sec t =
a2
a2
⋅ tan t ⋅ sec t −
⋅ ln sec t + tan t + C 1
2
2
在 Rt Δ ABC 中，可设 ∠ B = t ,| BC |= a ，
则 | AB |= x ,| AC |=
x2 − a2
x
, sec t =
a
a
∴ tan t =
∴
∫
x2 − a2
x 2 − a 2 dx = a 2 ∫ tan t d sec t
=
a2
⋅
2
x2 − a2 x a2
⋅ −
⋅ ln
a
a 2
x2 − a2 + x
+ C1
a
a2
⋅ ln x + x 2 − a 2 + C
2
π
2 .当 x < − a 时，可设 x = a ⋅ sect ( − < t < 0 ) 同理可证
2
x
a2
综合讨论 1 , 2 得：∫ x 2 − a 2 dx = =
x2 − a2 −
⋅ ln x + x 2 − a 2 + C
2
2
=
- 32 -
x
2
x2 − a2 −
54.
∫
( x 2 − a 2 ) 3 dx =
x
3
⋅ (2 x 2 − 5a 2 ) x 2 − a 2 + ⋅ a 4 ⋅ ln x + x 2 − a 2 + C
8
8
3
2
证明：∫ ( x − a ) dx = x ⋅ ( x − a ) − ∫ xd ( x − a )
2
2 3
2
2
2
2
(a > 0)
3
2
3
2
1
3
= x ⋅ ( x − a ) − ∫ x ⋅ ⋅ (2x) ⋅ ( x 2 − a 2 ) 2 d x
2
2
2
3
1
= x ⋅ ( x 2 − a 2 ) 2 − 3∫ x 2 ( x 2 − a 2 ) 2 d x
3
1
= x ⋅ ( x 2 − a 2 ) 2 − 3∫ ( x 2 − a 2 + a 2 )( x 2 − a 2 ) 2 d x
3
3
1
= x ⋅ ( x 2 − a 2 ) 2 − 3∫ ( x 2 − a 2 ) 2 d x − 3a 2 ∫ ( x 2 − a 2 ) 2 d x
3
1
x
3a 2
移项并整理得：∫ ( x − a ) dx = ⋅ ( x 2 − a 2 ) 2 −
(x2 − a 2 ) 2 d x
4
4 ∫
1
x 2
a2
2
2 2
2
又∫ ( x − a ) d x =
x −a −
⋅ ln x + x 2 − a 2 + C
(公式 53)
2
2
联立①②得：
2
2 3
①
②
3
∫
55.
x
3x 2
3
( x 2 − a 2 ) 3 dx = ( x 2 − a 2 ) 2 −
⋅ a ⋅ x 2 − a 2 + ⋅ a 4 ⋅ ln x + x 2 − a 2 + C
4
8
8
3
2
x
a x
3x 2
3
=( −
) x2 − a 2 −
⋅ a ⋅ x 2 − a 2 + ⋅ a 4 ⋅ ln x + x 2 − a 2 + C
4
4
8
8
x
3
= ⋅ (2 x 2 − 5a 2 ) x 2 − a 2 + ⋅ a 4 ⋅ ln x + x 2 − a 2 + C
8
8
1
(x 2 − a 2 )3 + C
( a > 0)
3
1
证明：∫ x x 2 − a 2 dx = ∫ x 2 − a 2 dx 2
2
1
1
= ∫ (x2 − a 2 ) 2 d (x 2 − a 2 )
2
1
1+
1
1
2
2
= ×
⋅ (x − a ) 2 + C
1
2
1+
2
1
=
(x2 − a 2 )3 + C
3
∫x
x 2 − a 2 dx =
- 33 -
56 .
2
2
2
∫ x x − a dx =
x
a4
⋅ (2 x 2 − a 2 ) x 2 − a 2 −
⋅ ln x +
8
8
x2 − a2 + C
(a > 0)
证明：被积函数 f ( x ) = x 2
x 2 − a 2 的定义域为 { x | x > a 或 x < − a}
π
1.当 x > a 时，可令 x = a ⋅ sect ( 0 < t < ) , 则 x 2 x 2 − a 2 = a 2 sec 2 t | a tant |
2
π
∵ 0 < t < , tan t > 0 , ∴ x 2 x 2 − a 2 = a 3 sec 2 t ⋅ tant
2
∴
∫x
2
x 2 − a 2 dx = ∫ a 3 sec 2 t ⋅ tant d ( a sect ) = a 4 ∫ sec 3 t d tan 2 t dt
=
a4
3
a4
3
a4
3
a4
3
a4
3
2
2
∫ sect ⋅ 3 ⋅ sec t ⋅ tan t dt =
a4
3
∫ sect d tan
3
①
t
a4
tan 3 t d sec t
3 ∫
a4
=
⋅ sec t ⋅ tan 3 t −
tan t ( sec 2 t − 1) d sec t
3 ∫
a4
a4
2
=
⋅ sec t ⋅ tan 3 t −
tan
t
⋅
sec
td
sec
t
+
tan t d sec t
3 ∫
3 ∫
a4
a4
2
3
=
⋅ sec t ⋅ tan 3 t −
tan
t
⋅
sec
t
dt
+
tan t d sec t
3 ∫
3 ∫
a4
a4
移项并整理得： a 4 ∫ sec 3 t d tan 2 t dt =
⋅ sec t ⋅ tan 3 t +
tan t d sec t
4
4 ∫
又 ∫ tant d sect = sect ⋅ tant − ∫ sect d tant = sect ⋅ tant − ∫ sec 3 tdt
=
⋅ sec t ⋅ tan 3 t −
②
= sect ⋅ tant − ∫ (1 + tan 2 t ) ⋅ sectdt
= sect ⋅ tant − ∫ sect dt − ∫ tan 2 t sect dt
= sect ⋅ tant − ∫ sect dt − ∫ tant d sect
1
1
⋅ sect ⋅ tant − ∫ sect dt
2
2
1
1
= ⋅ sect ⋅ tant − ln sect + tant + C 1
③
2
2
a4
a4
a4
将③式代入②式得： a 4 ∫ sec 3 t d tan 2 t dt =
tan 3 t ⋅ sect +
sect ⋅ tant −
ln sect + tant + C 1
4
8
8
移项并整理得： ∫ tant d sect =
在 Rt Δ ABC 中，设 ∠ B = t , | BC |= a ,则 | AB | = x , | AC | =
x2 − a2
x2 − a2
1
x
, sect =
=
a
cos t a
∴ tant =
∴ a 4 ∫ sec 3 t d tan 2 t dt =
a4 x x2 − a2
⋅ ⋅
⋅
4 a
a3
x2 − a2 +
a4
⋅
8
x2 − a2 x a4
x+
⋅ −
⋅ ln
a
a
8
x
a2 x
a4
⋅ (x 2 − a 2 ) ⋅ x 2 − a 2 +
⋅ x2 − a2 −
⋅ ln x +
4
8
8
x
a4
= ⋅ (2 x 2 − a 2 ) x 2 − a 2 −
⋅ ln x + x 2 − a 2 + C
8
8
2. 当 x < − a ,即 − x > a 时，令 μ = − x ,则 x = − μ
=
由讨论 1得：∫ x 2
x 2 − a 2 dx = − ∫ μ 2
x2 − a2 + C2
μ 2 − a 2 dμ
−μ
a4
⋅( 2μ2 − a2 ) μ2 − a2 +
⋅ ln − μ + μ 2 − a 2 + C
8
8
x
a4
将 μ = − x 代入上式得：∫ x 2 x 2 − a 2 dx = ⋅ ( 2 x 2 − a 2 ) x 2 − a 2 −
⋅ ln x + x 2 − a 2 + C
8
8
4
x
a
综合讨论 1,2 得：∫ x 2 x 2 − a 2 dx = ⋅ ( 2 x 2 − a 2 ) x 2 − a 2 −
⋅ ln x + x 2 − a 2 + C
8
8
=
- 34 -
x2 − a2
+ C1
a
57 .
∫
x2 − a2
dx =
x
x 2 − a 2 − a ⋅ arccos
证法 1：被积函数 f(x) =
a
+C
| x|
( a > 0)
x2 − a2
的定义域为
x
1 . 当 x > a 时 , 可设 x = a ⋅ sect
x2− a2
a ⋅ tant
则
=
,
x
a ⋅ sect
{ x | x > a 或 x < − a}
(0 < t <
π
),
2
dx = a ⋅ sect ⋅ tant d t
x2− a2
a ⋅ tant ⋅ a ⋅ sect ⋅ tant
dt = ∫ a ⋅ tan 2 tdt
∫ x dx = ∫
a ⋅ sect
2
sin t
1 − cos 2 t
1
= a∫
dt = a ∫
dt = a ∫
dt − ∫ dt
2
2
cos t
cos t
cos 2 t
= a ⋅ tant − a ⋅ t + C
a
a
∵ x = a ⋅ sect, ∴ cost = , ∴ t = arccos
x
x
∴
在 Rt Δ ABC 中，设 ∠ B = t ,| BC | = a ，则 | AB | = x , | AC |=
∴ tant =
∴
∫
| AC |
=
| BC |
x2 − a2
x2 − a2
a
x2− a2
dx = a ⋅ tant − a ⋅ t + C
x
a
+C
x
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
=
由讨论 1可知
∫
x 2 − a 2 − a ⋅ arccos
x2− a2
dx =
x
=
综合讨论 1 , 2 ，可写成： ∫
∫
μ2 − a2
dμ=
μ
μ 2 − a 2 − a ⋅ arccos
x 2 − a 2 − a ⋅ arccos
x2− a2
dx =
x
a
+C
μ
a
+C
−x
x 2 − a 2 − a ⋅ arccos
a
+C
| x|
- 35 -
57 .
∫
x2 − a2
dx =
x
x 2 − a 2 − a ⋅ arccos
a
+C
|x|
( a > 0)
x2 − a2
证法 2 ：被积函数 f(x) =
的定义域为 { x | x > a 或 x < − a }
x
1 . 当 x > a 时 , 可设 x = a ⋅ ch t ( 0 < t ) ，
x2 − a2
a ⋅ sht sht
=
=
, dx = a ⋅ sht dt
x
a ⋅ cht cht
则
∴
∫
x2 − a 2
dx =
x
sht
sh 2 t
⋅
a
⋅
sht
dt
=
a
∫ cht
∫ cht dt
ch 2 t − 1
cht
= a∫
dt = a ∫ chtdt − a ∫ 2 dt
cht
ch t
1
= a ∫ chtdt − a ∫
dsht
dx
公式 19 ：∫ 2
1 + sh 2 t
x + a2
= a ⋅ sht − a ⋅ arctan ( sht ) + C
提示 ： ch 2 t − sh 2 t = 1
( cht ) ′ = sht
( sh t ) ′ = cht
∵ x = a ⋅ cht, ∴ cht =
x
, ∴ sht = 1 − ch 2 t =
a
在 Rt Δ ABC 中，设 tany = sht =
1
x
⋅ arctan + C
a
a
x2 − a2
a
x2 − a2
, ∠ B = y ,| BC |= a
a
x 2 − a 2 , | AB |= | AC | 2 + | BC |2 = x
∴ y = arctan ( sht ), | AC |=
∴ cosy =
=
| BC | a
=
| AB | x
即 cosy = cos arctan ( sht ) =
∴ arctan ( sht ) = arccos
a
x
a
x
x2 − a2
a
∴ ∫
dx = x 2 − a 2 − a ⋅ arccos + C
x
x
2 .当 x < − a ,即 − x > a 时，令 μ = − x ,即 x = − μ
由讨论 1可知
∫
x2 − a2
dx =
x
=
综合讨论 1 , 2 ，可写成：∫
- 36 -
∫
μ2 − a2
dμ =
μ
x 2 − a 2 − a ⋅ arccos
x2 − a2
dx =
x
μ 2 − a 2 − a ⋅ arccos
a
+C
−x
x 2 − a 2 − a ⋅ arccos
a
+C
|x|
a
+C
μ
58.
∫
x2 − a 2
x2 − a2
dx
=
−
+ ln x + x 2 − a 2 + C
2
x
x
证明：∫
(a > 0)
x2 − a 2
1
dx = − ∫ x 2 − a 2 d
2
x
x
x2 − a2
1
=−
+ ∫ d x2 − a2
x
x
−
1
2
x2 − a2
1 1
=−
+ ∫ ⋅ ⋅ 2 x ⋅ ( x 2 − a 2 ) dx
x
x 2
x2 − a2
=−
+∫
x
1
x2 − a2
dx
公式 45：
∫
dx
2
x −a
2
= ln x +
x2 − a2 + C
x2 − a2
+ ln x + x 2 − a 2 + C
x
=−
（八）含有 a 2 − x 2 ( a > 0 ) 的积分（59~72）
59.
∫
dx
x
= arcsin + C
a
a2 − x2
(a > 0)
1
证明：被积函数 f ( x ) =
的定义域为 {x | −a < x < a}
a − x2
π
π
1
1
∴ 可设 x = a⋅ sint (− < t < ) ，则 dx = a ⋅ cos t dt ,
=
2
2
2
2
a ⋅ cos t
a −x
π
π
< t < , cos t > 0 ∴
2
2
∵−
∴
2
∫
dx
2
a −x
2
=∫
1
a2 − x2
=
1
a ⋅ cos t
1
⋅ a ⋅ cos t dt
a ⋅ cos t
= ∫ dt
= t +C
∵ x = a ⋅ sin t
∴
∫
∴ t = arcsin
x
a
dx
x
= arcsin + C
a
a2 − x 2
- 37 -
60.
dx
∫
(a 2 − x 2 )3
=
x
+C
a2 ⋅ a2 − x2
1
证明：被积函数 f ( x ) =
(a 2 − x 2 ) 3
∴ 可设 x = a⋅ sint (−
∵−
∴
∫
(a 2 − x 2 ) 3
=∫
的定义域为 {x | −a < x < a}
π
π
1
1
< t < ) ，则 dx = a ⋅ cos t dt ,
= 3
2
2
3
2
2
a ⋅ cos3 t
(a − x )
π
π
< t < , cos t > 0 ∴
2
2
dx
(a > 0)
1
2
2 3
(a − x )
=
1
a ⋅ cos 3 t
3
1
⋅ a ⋅ cos t dt
a ⋅ cos 3 t
3
1
dt
a ⋅ cos 2 t
1
= ∫ 2 ⋅ sec 2 t dt
a
1
= 2 ⋅ tan t + C
a
=∫
2
在Rt ΔABC中，设 ∠B = t ,| AB |= a ，
则| AC |= x ,| BC |= a 2 − x 2
x
∴ tan t =
2
a − x2
dx
x
∴ ∫
=
+C
(a 2 − x 2 )3 a 2 ⋅ a 2 − x 2
61.
∫
x
2
a −x
证明：∫
2
dx = − a 2 − x 2 + C
x
a2 − x2
( a > 0)
1
dx =
−
1
2
2
2
(
a
−
x
)
dx 2
∫
2
1
−
1
( a 2 − x 2 ) 2 d (a 2 − x 2 )
∫
2
1
1−
1
1
2
=− ×
⋅ ( a − x) 2 + C
2 1− 1
2
=−
= − a2 − x2 + C
- 38 -
62.
x
∫
2
2 3
dx =
(a − x )
1
+C
2
a − x2
( a > 0)
3
x
−
1
证明：∫
dx = ∫ (a 2 − x 2 ) 2 dx 2
2
(a 2 − x 2 ) 3
3
−
1
(a 2 − x 2 ) 2 d ( a 2 − x 2 )
∫
2
3
1−
1
1
2
2
=− ×
⋅ (a − x ) 2 + C
3
2
1−
2
1
=
+C
a2 − x2
=−
63 .
∫
x2
dx = −
a2 − x2
x
a2
x
a2 − x2 +
⋅ arcsin + C
2
2
a
(a > 0 )
x2
证明 ：被积函数 f ( x ) =
的定义域为 { x | − a < x < a}
a2 − x2
π
π
x2
a 2 ⋅ sin 2 t
∴ 可设 x = a ⋅ sint ( − < t < ) ， 则 dx = a ⋅ cos t dt ,
=
2
2
a ⋅ cost
a2 − x2
π
π
x2
a ⋅ sin 2 t
∵ − <t<
, cos t > 0 ∴
=
2
2
cost
a2 − x2
∫
∴
x2
a2 − x2
dx =
∫
a ⋅ sin 2 t
⋅ a ⋅ cos t dt
cost
提示：
cos2t = cos 2 t − sin 2 t
= a 2 ∫ sin 2 t dt
1 − cos 2t
dt
2
a2
dt
−
cos 2 t d ( 2 t )
∫
4 ∫
a2
⋅t −
⋅ sin2t + C
4
a2
⋅t −
⋅ sint ⋅ cost + C
2
= a2 ∫
a2
2
a2
=
2
a2
=
2
=
在 Rt Δ ABC 中，设 ∠ B = t , | AB | = a ，
则 | AC | = x , | BC | =
∴
∫
= 1 − 2sin 2 t
sin2t = 2 ⋅ sint ⋅ cost
a2 − x2
x
a2 − x2
∴ sint =
, cost =
a
a
2
x
x
a2
x
dx = −
a2 − x2 +
⋅ arcsin + C
2
2
2
2
a
a −x
- 39 -
64 .
∫
x2
2
2
(a − x )
3
dx =
x
2
a −x
x2
证明 ：被积函数 f ( x ) =
(a 2 − x 2 ) 3
∴ 可设 x = a ⋅ sint
∵ −
∴
∫
− arcsin
2
x
+C
a
的定义域为 { x | − a < x < a}
π
π
< t < ) ， 则 dx = a ⋅ cos t dt ,
2
2
(−
π
π
<t<
, cos t > 0 ∴
2
2
x2
a2 − x2
dx =
∫
(a > 0 )
x2
(a 2 − x 2 ) 3
=
x2
(a 2 − x 2 ) 3
=
sin 2 t
a ⋅ cos 3 t
sin 2 t
⋅ a ⋅ cos t dt
a ⋅ cos 3 t
sin 2 t
∫ cos 2 t dt
1 − cos 2 t
=∫
dt
cos 2 t
1
=∫
dt − ∫ dt
cos 2 t
= ∫ d tant − ∫ dt
=
= tant − t + C
在 Rt Δ ABC 中，设 ∠ B = t , | AB | = a ，
则 | AC | = x , | BC | =
∴ tant =
∴
- 40 -
∫
x2
(a 2 − x 2 ) 3
x
2
a − x2
dx =
x
a2 − x2
− arcsin
x
+C
a
a2 − x2
a 2 ⋅ sin 2 t
a 3 ⋅ cos 3 t
65 .
∫x
dx
a2 − x2
=
a2 − x2
+C
x
1
a−
⋅ ln
a
1
证明 ：被积函数 f ( x ) =
x a2 − x2
(a > 0 )
的定义域为 { x | − a < x < a 且 x ≠ 0}
1.当 − a < x < 0 时，可设 x = a ⋅ sint
(−
x a 2 − x 2 = a ⋅ sint ⋅ | a ⋅ cos t | ∵ −
∴
dx
∫x
2
a −x
=
2
∫a
1
a
1
=
a
=
2
π
< t < 0 ) ，则 dx = a ⋅ cos t dt
2
π
< t < 0 , cos t > 0 ∴ x a 2 − x 2 = a 2 ⋅ sint ⋅ cos t
2
1
⋅ a ⋅ cos t dt
⋅ sint ⋅ cos t
1
∫ sint
dt
sint
∫ sin
2
t
dt
1
1
d cos t
∫
a 1 − cos 2 t
1
1
1
−
(
+
)d cos t
∫
2 a 1 + cost 1 − cost
1
1
1
1
−
d (cos t + 1 ) +
d ( 1 − cos t )
∫
∫
2 a 1 + cost
2 a 1 − cost
1
1
−
⋅ ln 1 + cost +
⋅ ln cost − 1 + C 1
2a
2a
1
cost − 1
⋅ ln
+ C1
2a
1 + cost
=−
=
=
=
=
=
1
( cost − 1 ) 2
⋅ ln
⋅ ( − 1) + C 1
2a
1 − cos 2 t
=
1
( cost − 1 ) 2
⋅ ln
+ C2
2a
sin 2 t
1
cost − 1
⋅ ln
+ C2
a
sint
1
= ⋅ ln cott − csct + C 2
a
=
在 Rt Δ ABC 中，设 ∠ B = t ,| AB |= a ，
则 | AC |= x ,| BC |=
a2 − x2
1
a
, csct =
=
x
sint
x
∴ cott =
∴
∫x
a2 − x2
dx
1
⋅ ln
a
=
a2 − x2
=
a2 − x2 − a
1
a − a2 − x2
+ C 2 = ⋅ ln
⋅ ( − 1) + C 2
x
a
x
1
a − a2 − x2
⋅ ln
+ C3
a
x
∵ a − a2 − x2 > 0
∴
∫x
dx
a2 − x2
=
1
a − a2 − x2
⋅ ln
+C
a
x
2 .当0 < x < a 时，可设 x = a ⋅ sint
综合讨论 1 , 2 得：
∫
dx
x a2 − x2
=
(0 < t <
π
), 同理可证
2
1
a − a2 − x2
⋅ ln
+C
a
x
- 41 -
66 .
∫x
dx
2
=−
a2 − x2
a2 − x2
+C
a2x
证明 ：被积函数 f ( x ) =
(a > 0 )
1
x2 a2 − x2
的定义域为 { x | − a < x < a 且 x ≠ 0}
1.当 − a < x < 0 时，可设 x = a ⋅ sint
1
x
2
2
2
=
(−
π
< t < 0) ，则 dx = a ⋅ cos t dt ,
2
1
1
⋅
2
a ⋅ sin t a ⋅ cos t
2
a −x
π
π
1
1
∵ − <t<
, cos t > 0 ∴
= 3
2
2
a ⋅ sin 2 t ⋅ cost
x2 a2 − x2
dx
1
∴ ∫
=∫ 3
⋅ a ⋅ cos t dt
2
2
2
a ⋅ sin 2 t ⋅ cost
x a −x
=
1
a2
1
∫ sin
2
t
dt
1
− csc 2 t dt
2 ∫
a
1
= − 2 ⋅ cott + C
a
=−
在 Rt Δ ABC 中，设 ∠ B = t , | AB | = a ，
则 | AC | = x , | BC | =
a2 − x2
x
∴ cott =
∴
∫x
dx
2
a2 − x2
=−
a2 − x2
+C
a2x
2 .当0 < x < a 时，可设 x = a ⋅ sint
综合讨论 1 , 2 得：
∫
- 42 -
dx
x2 a2 − x2
(0 < t <
=−
π
), 同理可证
2
a2 − x2
+C
a2 x
a2 − x2
67 .
∫
a 2 − x 2 dx =
x
a2
x
a2 − x2 +
⋅ arcsin + C
2
2
a
证明 ：被积函数 f ( x ) =
∴ 可设 x = a ⋅ sint
∵ −
∫
∴
(a > 0 )
a 2 − x 2 的定义域为 { x | − a < x < a}
π
π
( − < t < ) ， 则 dx = a ⋅ cos t dt , a 2 − x 2 = a ⋅ cos t
2
2
π
π
<t<
, cos t > 0 ∴
2
2
a 2 − x 2 = a ⋅ cos t
a 2 − x 2 dx = ∫ a ⋅ cos t ⋅ a ⋅ cos t dt
= a 2 ∫ cos 2 t dt
= a 2 ∫ ( 1 − sin 2 t ) dt
= a 2 ∫ dt − a 2 ∫ sin 2 t dt
①
又 ∫ a 2 − x 2 dx = a 2 ∫ cos 2 t dt
= a 2 ∫ cost d sint
= a 2 ⋅ sint ⋅ cost − a 2 ∫ sint d cost
= a 2 ⋅ sint ⋅ cost + a 2 ∫ sin 2 t dt
②
由① + ②得：2 ∫ a 2 − x 2 dx = a 2 ∫ dt + a 2 ⋅ sint ⋅ cost = a 2 t + a 2 ⋅ sint ⋅ cost
∴
∫
a 2 − x 2 dx =
a2
a2
t+
⋅ sint ⋅ cost + C
2
2
在 Rt Δ ABC 中，设 ∠ B = t ,| AB |= a ，
则 | AC |= x ,| BC |=
∴ sint =
∴
∫
x
a
, cost =
a2 − x2
a2 − x2
a
a2
x a2
a2 − x2 x
⋅ arcsin +
⋅
⋅ +C
2
a
2
a
a
2
x
a
x
=
a2 − x2 +
⋅ arcsin + C
2
2
a
a 2 − x 2 dx =
- 43 -
68.
∫
( a 2 − x 2 ) 3 dx =
x
3
x
⋅ (5a 2 − 2 x 2 ) a 2 − x 2 + ⋅ a 4 ⋅ arcsin + C
8
8
a
3
2
证明：∫ (a − x ) dx = x ⋅ ( a − x ) − ∫ xd ( a − x )
2
2 3
2
2
2
2
(a > 0)
3
2
3
1
3
= x ⋅ ( a 2 − x 2 ) 2 − ∫ x ⋅ ⋅ ( −2 x) ⋅ ( a 2 − x 2 ) 2 d x
2
3
1
= x ⋅ ( a 2 − x 2 ) 2 + 3∫ x 2 ( a 2 − x 2 ) 2 d x
2
2
3
2
2
2
2
2
3
2
2
2
2
2
1
2
2
= x ⋅ (a − x ) + 3∫ ( x − a + a )( a − x ) d x
3
2
= x ⋅ (a − x ) − 3∫ ( a − x ) d x + 3a
2
∫ (a
2
2
1
2
−x ) d x
3
1
x
3a 2
移项并整理得：∫ ( a − x ) dx = ⋅ ( a 2 − x 2 ) 2 +
(a 2 − x 2 ) 2 d x
∫
4
4
1
2
x 2
a
x
又∫ ( a 2 − x 2 ) 2 d x =
a − x2 +
⋅ arcsin + C
(公式 67)
2
2
a
联立①②得：
2
2 3
①
②
3
∫
69 .
∫x
x
3x 2
3
x
( a − x ) dx = (a 2 − x 2 ) 2 +
⋅ a ⋅ a 2 − x 2 + ⋅ a 4 ⋅ arcsin + C
4
8
8
a
2
3
a x x
3x 2
3
x
=(
− ) a2 − x2 +
⋅ a ⋅ a 2 − x 2 + ⋅ a 4 ⋅ arcsin + C
4
4
8
8
a
x
3
x
= ⋅ (5a 2 − 2 x 2 ) a 2 − x 2 + ⋅ a 4 ⋅ arcsin + C
8
8
a
2
2 3
a 2 − x 2 dx = −
1
(a 2 − x 2 ) 3 + C
3
(a > 0)
证明 ：被积函数
f ( x ) = x a 2 − x 2 的定义域为 { x | − a < x < a}
π
π
∴ 可设 x = a ⋅ sint ( − < t < ) ， 则 dx = a ⋅ cos t dt , x a 2 − x 2 = a ⋅ sin t ⋅ | a ⋅ cos t |
2
2
π
π
∵ − <t<
, cos t > 0 ∴ x a 2 − x 2 = a 2 ⋅ sint ⋅ cost
2
2
∴
∫x
a 2 − x 2 dx = ∫ a 2 ⋅ sint ⋅ cost ⋅ a ⋅ cos t dt = a 3 ∫ cos 2 t ⋅ sint dt
= − a 3 ∫ cos 2 t dcost = −
a3
cos 3 t + C
3
3
a3
=−
(1 − sin 2 t ) 2 + C
3
π
π
x
( − < t < ) , ∴ sint =
2
2
a
∵ x = a ⋅ sint
3
3
(a 2 − x 2 )3
a2 − x2 2
)
=
a2
a3
3
a3
2
2
2
∴ ∫ x a − x dx = −
(1 − sin t ) 2 + C
3
1
=−
(a 2 − x 2 )3 + C
3
∴ (1 − sin 2 t ) 2 = (
- 44 -
70 .
2
2
2
∫ x a − x dx =
x
a4
x
⋅ (2 x 2 − a 2 ) a 2 − x 2 +
⋅ arcsin + C
8
8
a
(a > 0)
证明：被积函数 f ( x ) = x 2 a 2 − x 2 的定义域为 { x | − a < x < a}
∴ 可令 x = a ⋅ sint
∵ −
∴
(−
π
π
< t < ),
2
2
则 x 2 a 2 − x 2 = a 2 ⋅ sin 2 t | a ⋅ cost |
π
π
< t < , cos t > 0 , ∴ x 2 a 2 − x 2 = a 3 ⋅ sin 2 t ⋅ cost
2
2
∫x
2
a 2 − x 2 dx = ∫ a 3 sin 2 t ⋅ cost d ( a ⋅ sint ) = a 4 ∫ sin 2 t ⋅ cos 2 t dt
=
=
=
=
=
=
移项并整理得：
∫x
a4
3
a4
3
a4
3
a4
3
a4
3
a4
3
2
∫ 3 ⋅ sin
2
t ⋅ cos t ⋅ cos t dt
∫ cos t d sin
3
t
a4
3
a4
⋅ cos t ⋅ sin 3 t −
3
a4
⋅ cos t ⋅ sin 3 t −
3
a4
⋅ cos t ⋅ sin 3 t −
3
⋅ cos t ⋅ sin 3 t −
∫ sin
3
t d cost
∫ sint ⋅ (1 − cos
2
t ) d cost
a4
3
a4
sint
d
cost
+
∫
3
∫ sint d cost +
∫ sint ⋅ cos
∫ sin
2
2
t d cost
t ⋅ cos 2 t dt
a 2 − x 2 dx = a 4 ∫ sin 2 t ⋅ cos 2 t dt
a4
a4
⋅ cos t ⋅ sin 3 t −
4
4
∵ ∫ sint d cost = sint ⋅ cost − ∫ cost d sint
=
∫ sint d cost
①
= sint ⋅ cost − ∫ cos 2 t dt
= sint ⋅ cost − ∫ (1 − sin 2 t )dt
= sint ⋅ cost − ∫ dt + ∫ sin 2 t dt
②
又 ∵ ∫ sint d cost = − ∫ sin 2 t dt
③
1
1
⋅ sint ⋅ cost − ∫ dt
2
2
1
1
= ⋅ sint ⋅ cost − ⋅ t + C 1
2
2
4
a
a4
a4
a 2 − x 2 dx =
⋅ cos t ⋅ sin 3 t −
⋅ sint ⋅ cost −
⋅t + C
4
8
8
联立②③得：∫ sint d cost =
联立①④得：∫ x 2
在 Rt Δ ABC 中，设 ∠ B = t , | A B | = a , 则 | AC | = x , | BC | =
∴ cost =
∴
④
a2 − x2
a2 − x2
x
, sint =
a
a
a4
a 2 − x 2 x3 a 4
a2 − x2 x a4
x
⋅
⋅ 3 −
⋅
⋅ +
⋅ arcsin + C
4
a
8
a
a
8
a
a
4
x
a
x
= ⋅ (2 x 2 − a 2 ) a 2 − x 2 +
⋅ arcsin + C
8
8
a
2
2
2
∫ x a − x dx =
- 45 -
71 .
∫
a2 − x2
dx =
x
a 2 − x 2 + a ⋅ ln
证明 ：被积函数 f ( x ) =
a − a2 − x2
+C
x
1
x a2 − x2
(a > 0 )
的定义域为 { x | − a < x < a 且 x ≠ 0}
1.当 − a < x < 0 时，可设 x = a ⋅ sint
(−
π
< t < 0) ，则 dx = a ⋅ cos t dt
2
a2 − x2
| a ⋅ cos t |
π
=
∵ − < t < 0 , cos t > 0 ∴
x
a ⋅ sint
2
∴
a2 − x2
cos t
=
x
sint
a2 − x2
dx =
x
cos t
cos 2 t
⋅
a
⋅
cos
t
dt
=
a
∫ sint
∫ sint dt
1 − sin 2 t
1
= a∫
dt = a ∫
dt − a ∫ sint dt
sint
sint
sint
1
= a∫
dt − a ∫ sint dt = − a ∫
dcost − a ∫ sint dt
2
sin t
1 − cos 2 t
a
1
1
= − ∫(
+
) d cos t − a ∫ sint dt
2 1 + cost 1 − cost
a
1
a
1
=− ∫
d (cos t + 1) + ∫
d (cost − 1) − a ∫ sint dt
2 1 + cost
2 cost − 1
a
a
= − ⋅ ln 1 + cost + ⋅ ln cost − 1 + a ⋅ cos t + C 1
2
2
a
cost − 1
= ⋅ ln
+ a ⋅ cos t + C 1
2
1 + cost
∫
=
a
( cost − 1) 2
⋅ ln
⋅ ( − 1) + a ⋅ cos t + C 1
2
1 − cos 2 t
=
a
( cost − 1) 2
⋅ ln
+ a ⋅ cos t + C 2
2
sin 2 t
= a ⋅ ln
cost − 1
+ a ⋅ cos t + C 2
sint
= a ⋅ ln cott − csct + a ⋅ cos t + C 2
在 Rt Δ ABC 中，设 ∠ B = t , | AB | = a ，
则 | AC | = x , | BC | =
∴ cott =
∴
a2 − x2
1
a
, csct =
=
, cost =
x
sint
x
a2 − x2
dx = a ⋅ ln
x
∫
= a ⋅ ln
=
a2 − x2 − a
+ a⋅
x
+ a⋅
a2 − x2
a
a2 − x2
+ C2
a
a − a2 − x2
⋅ ( − 1) + a ⋅
x
1
a − a2 − x2
⋅ ln
a
x
a2 − x2
a2 − x2
+ C2
a
a2 − x2
+ C3
a
∵ a − a2 − x2 > 0
∴
∫
a2 − x2
dx =
x
a 2 − x 2 + a ⋅ ln
2 .当0 < x < a 时，可设 x = a ⋅ sint
综合讨论 1 , 2 得：∫
- 46 -
a2 − x2
dx =
x
a − a2 − x2
+C
x
(0 < t <
π
), 同理可证
2
a 2 − x 2 + a ⋅ ln
a − a2 − x2
+C
x
72 .
∫
a2 − x2
a2 − x2
x
dx
=
−
− arcsin + C
2
x
a
x
证明 ：被积函数 f ( x ) =
a2 − x2
的定义域为 { x | − a < x < a 且 x ≠ 0}
x2
1.当 − a < x < 0 时，可设 x = a ⋅ sint
∵ −
∴
∫
(a > 0 )
π
< t < 0 , cos t > 0 ∴
2
(−
π
< t < 0 ) ， 则 dx = a ⋅ cos t dt ,
2
a ⋅ cos t
a2 − x2
= 2
2
x
a ⋅ sin 2 t
a2 − x2
cos t
=
2
x
a ⋅ sin 2 t
a2 − x2
cos t
dx = ∫
⋅ a ⋅ cos t dt
2
x
a ⋅ sin 2 t
cos 2 t
=∫
dt
sin 2 t
1 − sin 2 t
=∫
dt
sin 2 t
= ∫ csc 2 tdt − ∫ dt
= − cott − t + C
在 Rt Δ ABC 中，设 ∠ B = t , | AB | = a ，
则 | AC | = x , | BC | =
∴ cott =
a2 − x2
a2 − x2
x
a2 − x2
x
− arcsin + C
∫
x
a
π
2 .当0 < x < a 时，可设 x = a ⋅ sint ( 0 < t < ), 同理可证
2
∴
a2 − x2
dx = −
x2
综合讨论 1 , 2 得：
∫
a2 − x2
dx = −
x2
a2 − x2
x
− arcsin + C
x
a
- 47 -
（九）含有 ± a 2 + bx + c ( a > 0) 的积分（73~78）
73 .
dx
1
⋅ ln 2 ax + b + 2 a ax 2 + bx + c + C
( a > 0)
2
a
ax + bx + c
1
证明：
若被积函数 f(x) =
成立，则ax 2 + bx + c > 0恒成立
2
ax + bx + c
∵ a > 0 ∴ Δ = b 2 − 4ac > 0
1
∵ ax 2 + bx + c =
[(2 ax + b ) 2 + 4 ac − b 2 ]
4a
1
=
[(2 ax + b) 2 − ( b 2 − 4ac ) 2 ]
4a
dx
1
∴ ∫
= 2 a∫
dx
2
ax + bx + c
( 2ax + b) 2 − ( b 2 − 4ac ) 2
∫
=
=
=
2 a
2a
1
a
∫
∫
1
d ( 2ax + b)
2
2
( 2ax + b) − ( b − 4ac )
1
2
2
d (2 ax + b )
2
(2 ax + b ) − ( b − 4ac )
2
公式45：∫
dx
x2 − a 2
= ln | x + x 2 − a 2 | + C
1
⋅ ln 2 ax + b + ( 2ax + b) 2 − ( b 2 − 4ac ) 2 + C
a
1
=
⋅ ln 2 ax + b + 4a ⋅ ( ax 2 + bx + c) + C
a
1
=
⋅ ln 2 ax + b + 2 a ax 2 + bx + c + C
a
=
74.
ax 2 + bx + c dx =
∫
2 ax + b
4ac − b 2
ax 2 + bx + c +
⋅ ln 2 ax + b + 2 a ax 2 + bx + c + C
3
4a
8 a
( a > 0)
证明：
若被积函数 f(x) = ax 2 + bx + c成立，则ax 2 + bx + c > 0恒成立
∵a>0
∴ Δ = b 2 − 4ac > 0
1
[(2 ax + b ) 2 + 4ac − b 2 ]
4a
1
=
[(2 ax + b ) 2 − ( b 2 − 4ac ) 2 ] 公式53：∫ x2 − a2 dx = x
2
4a
1
ax 2 + bx + c dx =
( 2ax + b) 2 − ( b 2 − 4ac ) 2 dx
2 a∫
1
=
(2 ax + b ) 2 − ( b 2 − 4ac ) 2 d ( 2ax + b)
∫
2a ⋅ 2 a
∵ ax 2 + bx + c =
∴
∫
x2 − a 2 −
a2
⋅ ln x+ x 2 − a 2 + C
2
⎡ 2 ax + b
⎤
b 2 − 4ac
⋅⎢
(2 ax + b ) 2 − ( b 2 − 4ac ) 2 −
⋅ ln 2 ax + b + ( 2ax + b) 2 − ( b 2 − 4ac ) 2 ⎥
2
4a ⋅ a ⎣ 2
⎦
2
1
2 ax + b
4ac − b
=
⋅
⋅ 2 a ax 2 + bx + c +
⋅ ln 2ax + b + (2 ax + b ) 2 − ( b 2 − 4ac ) 2 + C
3
3
2
4 a
8 a
1
2 ax + b
4ac − b 2
=
⋅
⋅ 2 a ax 2 + bx + c +
⋅ ln 2ax + b + 4 a ⋅ (ax 2 + bx + c) + C
3
3
2
4 a
8 a
1
=
=
2ax + b
4ac − b 2
⋅ ax 2 + bx + c +
⋅ ln 2ax + b + 2 a ax 2 + bx + c + C
3
4a
8 a
- 48 -
75.
x
1
b
2
ax
+
bx
+
c
−
⋅ ln 2ax + b + 2 a ax2 + bx + c + C
∫ ax2 + bx + c a
3
2 a
2
证明：∵ d (ax + bx + c) = (2ax + b)dx
dx =
(a > 0)
⎡
1
⎛ 2ax + b b ⎞⎤
dx 变换成∫ ⎢
⋅⎜
− ⎟⎥ dx
2
2
a
2a ⎠⎦
⎝
ax2 + bx + c
ax
+
bx
+
c
⎣
1
1
b
1
∴ 上式 = ∫
⋅ (2ax + b)dx − ∫
dx
2
2
2a ax + bx + c
2a ax + bx + c
1
−
1
b
1
= ∫ (ax2 + bx + c ) 2 d (ax2 + bx + c) − ∫
dx
2a
2a ax2 + bx + c
x
∴ 可将∫
1
b
1
ax2 + bx + c − ∫
dx
a
2a ax2 + bx + c
1
b 1
dx = ⋅
⋅ ln 2ax + b + 2 a ax2 + bx + c + C1
2
2a a
ax + bx + c
=
又
b
2a ∫
=
∴
76.
∫
∫
x
ax2 + bx + c
dx
c + bx − ax
2
=
1
a
dx =
b
2 a3
(公式73)
⋅ ln 2ax + b + 2 a ax2 + bx + c + C1
1
b
ax2 + bx + c −
⋅ ln 2ax + b + 2 a ax2 + bx + c + C
3
a
2 a
2ax − b
⋅ arcsin
b 2 + 4ac
1
证明：
若被积函数 f(x) =
2
+C
(a > 0)
成立，则c + bx − ax 2 > 0有解
c + bx − ax
∵ a > 0 ∴ Δ = b 2 + 4ac > 0
1
∵ c + bx − ax 2 = [b 2 − (2ax − b) 2 ] + c
4a
b 2 + 4ac (2ax − b) 2
=
−
4a
4a
dx
1
∴ ∫
= 2 a∫
dx
c + bx − ax 2
(b 2 + 4ac) 2 − (2ax − b) 2
=
原题：
∫
dx
c + bx − ax 2
=−
1
a
1
a
⋅ arcsin
⋅ arcisn
2ax − b
b 2 + 4ac
+C
2 ax − b
b 2 + 4ac
+ C 有误
- 49 -
77.
∫
2ax − b
b 2 + 4ac
2ax − b
2
c + bx − ax dx =
c + bx − ax +
⋅ arcsin
+C
3
8a
8 a
b 2 + 4ac
2
( a > 0)
证明：
若被积函数 f(x) = c + bx − ax 2 成立，则c + bx − ax 2 ≥ 0有解
∵ a >0
∴ Δ = b 2 + 4ac ≥ 0
1 2
[b − (2ax − b) 2 ] + c
4a
b 2 + 4ac (2ax − b) 2
=
−
公式 67：∫ a 2 − x 2 dx =
4a
4a
1
c + bx − ax 2 dx =
(b 2 + 4ac) 2 − ( 2ax − b) 2 dx
∫
2 a
1
=
( b 2 + 4ac ) 2 − (2ax − b) 2 d ( 2ax − b)
∫
2 a ⋅ 2a
∵ c + bx − ax 2 =
∴
∫
x
a2
x
a2 − x2 +
⋅ arcsin + C
2
2
a
⎡ 2ax − b
b 2 + 4ac
2ax − b ⎤
2
2
2
(
b
+
4ac
)
−
(
2
ax
−
b
)
+
⋅ arcsin
⎢
⎥+C
2
4 a3 ⎣ 2
b 2 + 4ac ⎦
2ax − b
b 2 + 4ac
2ax − b
=
4a ⋅ (c + bx − ax 2 ) +
⋅ arcsin
+C
3
3
8 a
8 a
b 2 + 4ac
2ax − b
b 2 + 4ac
2ax − b
=
c + bx − ax 2 +
⋅ arcsin
+C
8a
8 a3
b 2 + 4ac
=
78.
x
1
1
b
2ax − b
c + bx − ax 2 +
⋅ arcsin
+C
3
a
c + bx − ax
2 a
b 2 + 4ac
x
证明：
若被积函数 f(x) =
成立，则c + bx − ax 2 > 0有解
2
c + bx − ax
∵ a > 0 ∴ Δ = b 2 + 4ac > 0
1
∵ c + bx − ax2 = [b 2 − (2ax − b) 2 ] + c
4a
1 2
=
b + 4ac − (2ax − b) 2
4a
x
x
∴ ∫
dx = 2 a ∫
dx 公式 61 ：∫
2
2
2
c + bx − ax2
( b + 4ac ) − (2ax − b)
∫
2
dx = −
[
=2 a⋅
2 a
=−
=−
]
3
∫
1
2 a
1
3
2ax − b
d (2ax − b) +
2 a
( b 2 + 4ac) 2 − (2ax − b) 2
( b 2 + 4ac) 2 − (2ax − b) 2 +
b
2 a
4a ⋅ (c + bx − ax2 ) +
b
b
3
⋅ arcsin
⋅ arcsin
3
∫
a
− x2
dx = −
1
a2 − x2 + C
2
b + 4ac
2ax − b
+C
d (2ax − b)
( b 2 + 4ac) 2 − (2ax − b) 2
2ax − b
2 a3
2 a3
b 2 + 4ac
1
b
2ax − b
=−
c + bx − ax2 +
⋅ arcsin
+C
a
2 a3
b 2 + 4ac
- 50 -
x
2
1 1
2ax − b + b
⋅ ∫
d (2ax − b)
2a 2a ( b 2 + 4ac ) 2 − (2ax − b) 2
1
=
(a > 0)
+C
公式59 : ∫
dx
x
= arcsin + C
a
a −x
2
2
（十）含有
79 .
∫
±
x − a
x − b
或
( x − a )( b − x ) 的积分（79~82）
x−a
x−a
dx = ( x − b )
+ (b − a ) ⋅ ln ( x − a +
x−b
x −b
x−a
> 0 可令t =
x−b
证明 : ∵
∫
∴
x−a
x −b
x−b)+C
(t > 0) ，则 x =
a − bt 2
2t ⋅ ( a − b )
，dx =
dt
2
1− t
(1 − t 2 ) 2
x−a
2t ⋅ ( a − b )
t2
dx = ∫ t ⋅
dt
=
2
(
a
−
b
)
∫ (1 − t 2 ) 2 dt
x−b
(1 − t 2 ) 2
1− t2 +1
1
1
dt = 2 (b − a ) ∫ [
−
]dt
2 2
2
(1 − t )
1− t
(1 − t 2 ) 2
1
1
1
1
= 2 (b − a ) ∫
dt − 2 (b − a ) ∫
dt = 2 ( a − b ) ∫ 2 dt + 2 ( a − b ) ∫
dt
2
2 2
1− t
(1 − t )
t −1
(1 − t 2 ) 2
= 2(b − a ) ∫
1
t −1
1
t −1
1
⋅ ln
+ 2( a − b) ∫
dt = ( a − b ) ⋅ ln
+ 2( a − b ) ∫
dt
2 2
2
t +1
t +1
(1 − t )
(1 − t 2 ) 2
1
1
对于 ∫
dt = ∫ 2
dt
(t > 0 )
2 2
(1 − t )
(t − 1) 2
π
∴ 可令 t = sec k
(0 < k < )，则 (t 2 − 1) 2 = tan 4 k , d sec k = sec k ⋅ tan kdk
2
1
1
sec k
cos 2 k
∴ ∫ 2
dt
=
⋅
sec
k
⋅
tan
kdk
=
dk
=
∫ tan 4 k
∫ tan 4 k
∫ sin 3 k dk
(t − 1) 2
= 2( a − b ) ⋅
1 − sin 2 k
1
1
1 cos k 1
1
1
dk = ∫
dk − ∫
dk = − ⋅
+ ∫
dk − ∫
dk
3
3
2
sin k
sin k
sin k
2 sin k 2 sin k
sin k
1 cos k 1
1
1
1 cos k
=− ⋅
− ∫
dk = − ⋅ ln csck − cotk − ⋅
2
2 sin k 2 sin k
2
2 sin 2 k
=∫
在 Rt Δ ABC中， ∠ B = k ，| BC | = 1 则 | AC | = t 2 − 1 ，| AB | = t
t
1
∴ csc k =
=
sin k
∴
∫
t 2 −1
1
, cos k = , sin k =
2
t
t −1
x−a
t −1
1
dx = ( a − b ) ⋅ ln
+ 2( a − b )[− ⋅ ln
x−b
t +1
2
= ( a − b ) ⋅ ln
= ( a − b ) ⋅ ln
将t =
, cot k =
x−a
代入上式得：∴
x−b
∫
t2 −1
t
1
t −1
− ( a − b ) ⋅ ln
t +1
t −1
t2 −1
t −1
t2 −1
−
−
t
2
2 (t − 1)
] + C1
( a − b) ⋅ t
+ C1
t2 −1
t 2 −1
( a − b) ⋅ t
− 2
+ C1
t +1
(t − 1)
x−a
dx = (a − b ) ⋅ ln
x−b
b−a
x−b
x−a +
− ( a − b)
x−b
x−a x−b
⋅
+ C1
x−b b−a
x−b
= ( x − b)
x−a
+ (a − b ) ln
x −b
b−a
x−a +
= ( x − b)
x−a
+ (a − b ) ln
x −b
= ( x − b)
x−a
+ (b − a ) ⋅ ln ( x − a +
x−b
x−b
+ C1
b − a + (b − a ) ln
x−b )+C
x−a +
x − b + C1
- 51 -
80.
∫
x−a
x−a
x−a
dx = ( x − b)
+ (b − a ) ⋅ arcsin
+C
b− x
b− x
b−a
x−a
x−a
> 0 可令t =
b− x
b−x
证明 : ∵
∴
∫
(t > 0) ，则x =
a + bt 2
2t ⋅ (b − a )
，dx =
dt
2
1+ t
(1 + t 2 ) 2
x−a
2t ⋅ (b − a)
t2
dx = ∫ t ⋅
dt
=
2
(
b
−
a
)
∫ (1 + t 2 ) 2 dt
b−x
(1 + t 2 ) 2
1+ t 2 −1
1
1
dt = 2(b − a )∫ [
−
]dt
2 2
2
(1 + t )
1+ t
(1 + t 2 ) 2
1
1
1
= 2 (b − a ) ∫
dt − 2(b − a) ∫
dt = 2(b − a)arcsint − 2(a − b)∫
dt
2
2 2
1+ t
(1 + t )
(1 + t 2 ) 2
= 2(b − a) ∫
∴
∫
1
t −1
1
t −1
1
= 2( a − b) ⋅ ⋅ ln
+ 2( a − b) ∫
dt = (a − b) ⋅ ln
+ 2( a − b) ∫
dt
2 2
2
t +1
(1 − t )
t +1
(1 − t 2 ) 2
1
对于 ∫
dt (t > 0)
(1 + t 2 ) 2
π
∴ 可令 t = tan k (0 < k < )，则 (t 2 + 1) 2 = sec 4 k , dt = sec 2 kdk
2
1
1
1
∴ ∫
dt = ∫
⋅ sec 2 kdk = ∫
dk = ∫ cos 2 kdk
2 2
4
2
(1 + t )
sec k
sec k
1
1
1
= ∫ (1 + cos 2k )dk = ∫ d k + ∫ cos 2kdk
2
2
2
k 1
= + ⋅ sin 2k + C1
2 4
x−a
k 1
dx = 2(b − a )k − 2(b − a )[ + ⋅ sin 2k ] + C1
b−x
2 4
= (b − a ) k − (b − a) sin k ⋅ cos k + C1
在 RtΔABC中， ∠ B = k ，| BC | = 1 则 | AC | = 1 ，| AB | = t 2 + 1
1
t
∴ cos k =
, sin k =
t2 +1
t2 +1
x−a
t
1
t
dx = (b − a) arcsin
− (b − a ) ⋅
⋅
+ C1
2
2
2
b− x
t +1
t +1 t + 1
t
t
= (b − a) arcsin
− (b − a ) ⋅ 2
+ C1
t +1
t2 +1
⎛ b−x x−a ⎞
x−a
x−a
x−a b−x
⎟ − (b − a ) ⋅
代入上式得：∴ ∫
dx = (b − a) arcsin ⎜⎜
+ C1
⎟
b−x
b−x
b−x b−a
⎝ b−a b−x ⎠
∴
将t =
- 52 -
∫
= (b − a ) arcsin
x−a
x−a
− (b − x ) ⋅
+ C1
b−a
b− x
= (b − a ) arcsin
x−a
x−a
+ ( x − b) ⋅
+C
b−a
b−x
81.
∫
dx
x−a
= 2 arcsin
+C
b−a
( x − a )( b − x )
证明 :
( a < b)
dx
1
x−a
=∫
⋅
dx
| x−a| b− x
( x − a)(b − x)
∫
令 t=
x−a
a + bt 2
(b − a )t 2
2t (b − a)
，则x =
，
|
x
−
a
|
=
，dx =
dt
2
2
b−x
1+ t
1+ t
(1 + t 2 ) 2
t2
1+ t 2
1
x−a
1 1+ t 2
2t ⋅ (b − a)
于是 ∫
⋅
dx = ∫
⋅ 2 ⋅t ⋅
dt
| x−a| b−x
b−a t
(1 + t 2 ) 2
1
= 2∫
dt =2 arctan t + C
(公式 19)
1+ t2
x−a
= 2 arctan
+C
b− x
∵ b > a , ∴ | x − a | = (b − a ) ⋅
令 tan μ =
x−a
x−a
，则 μ = arctan
b− x
b− x
在 RtΔABC中， ∠ B = μ ，| AC | = x − a
∴ | BC | = b − x ，| AB | = | AC | 2 + | BC | 2 = b − a
∴ sin μ =
∴
∫
x−a
x−a
, ∴ μ = arcsin
b−a
b−a
dx
x−a
= 2 arcsin
+C
b−a
( x − a )(b − x)
- 53 -
82.
∫
( x − a)(b − x) dx =
证明 :
∫
( x − a)(b − x) dx = ∫ x − a
b−x
> 0 可令t =
x−a
∵
2x − a − b
(b − a ) 2
x−a
( x − a )(b − x) +
⋅ arcsin
+ C ( a < b)
4
4
b−x
b−x
x−a
b−x
dx
x−a
(t > 0) ，则x =
b + at 2
2at ⋅ (1 + t 2 ) − 2t ( at 2 + b)
2t ( a − b)
，
dx
=
dt =
dt
2
2 2
1+ t
(1 + t )
(1 + t 2 ) 2
at 2 + b − a − at 2
b−a
x−a =
=
2
1+ t
1+ t 2
b−a
1+ t 2
b − a 2t (a − b)
( x − a )(b − x ) dx = ∫
⋅t ⋅
dt
1+ t2
(1 + t 2 ) 2
∵ a<b
∴
∫
∴ x−a =
= −2( a − b) 2 ∫
对于 ∫
t2
dt
(1 + t 2 ) 3
t2
dt
(1 + t 2 ) 3
(t > 0) ∴ 可令 t = tan k
π
(0 < k < )，则 (t 2 + 1) 3 = sec 6 k , dt = sec 2 kdk
2
t2
tan 2 k
tan 2 k
2
2
2
dt
=
⋅
sec
kdk
=
∫ (1 + t 2 ) 3
∫ sec 6 k
∫ sec 4 k dk = ∫ sin k ⋅ cos kdk
2
1
1
= ∫ ( 2 sin k ⋅ cos k ) dk = ∫ sin 2 2k ⋅dk
4
4
1 ⎡ 2k 1
⎤
= ⎢ − ⋅ sin 4k ⎥ + C
8⎣ 2 4
⎦
k 1
= − ⋅ sin 4k + C
8 32
k 1
= − ⋅ (4 sin k ⋅ cos 3 k − 4 sin 3 k ⋅ cos k ) + C
8 32
k 1
1
= − ⋅ sin k ⋅ cos 3 k + sin 3 k ⋅ cos k + C
8 8
8
1
联立以上两式得：∫ ( x − a )(b − x) dx = −2(b − a) 2 ⋅ ⋅ ( k − sin k ⋅ cos 3 k + sin 3 k ⋅ cos k ) + C
8
2
(b − a )
=−
⋅ ( k − sin k ⋅ cos 3 k + sin 3 k ⋅ cos k ) + C
4
∴
在 RtΔABC中， ∠ B = k ，| BC | = 1 则 | AC | = t ，| AB | = t 2 + 1
t
1
∴ cos k =
, sin k =
t2 +1
t2 +1
(b − a ) 2
t
1
t2
t
∴ ∫ ( x − a)(b − x) dx = −
⋅ ( arcsin
−
⋅ 2
⋅
4
t2 +1
t 2 +1 t +1 t 2 + 1
1
1
t
+ 2
⋅
)
t +1 t 2 +1 t 2 + 1
(b − a ) 2
t
t3
t
⋅ ( arcsin
− 2
+ 2
)+C
2
4
(t + 1) 2
t 2 + 1 (t + 1)
(b − a ) 2
t
t (t 2 − 1)
=−
⋅ ( arcsin
− 2
)+C
2
4
t 2 + 1 (t + 1)
=−
将t =
- 54 -
b−x
2x − a − b
(b − a ) 2
x−a
代入上式得：∫ ( x − a )(b − x) dx =
( x − a )(b − x) +
⋅ arcsin
+C
x−a
4
4
b−x
（十一）含有三角函数的积分（83~112）
83.
∫ sinx dx = −cosx + C
证明：∫ sinx dx = −∫ ( − sinx) dx
∵ (cosx) ′ = − sinx 即 cosx为 − sinx的原函数
∴ ∫ sinx dx = − ∫ dcosx
= −cosx + C
84.
∫ cos x dx = sin x + C
证明：∵ ( sin x ) ′ = cos x 即 sin x为cos x的原函数
∴ ∫ cos x dx = ∫ d sin x
= sin x + C
85.
∫ tan x dx = − ln
cosx + C
sinx
dx
cos x
1
= −∫
d cos x
cos x
= − ln cosx + C
证明：∫ tan x dx = ∫
86.
∫ cot x dx = ln
sin x + C
cos x
dx
sin x
1
=∫
d sin x
sin x
= ln sin x + C
证明：∫ cot x dx = ∫
87.
π
x
∫ sec xdx = ln | tan ( 4 + 2 ) | +C = ln | sec x + tan x| + C
1
cos x
dx = ∫
dx
cos x
cos 2 x
1
1
1
1
1
=∫
d sin x = ∫
d sin x + ∫
d sin x
2
1 − sin x
2 1 + sin x
2 1 − sin x
1
1
= ⋅ ln | 1 + sin x | − ⋅ ln | 1 − sin x | +C
2
2
证明：∫ sec xdx = ∫
(1 + sin x ) + C
1
1 + sin x
1
= ⋅ ln |
| +C = ⋅ ln
2
1 − sin x
2
1 − sin 2 x
2
1
(1 + sin x ) + C = ln | 1 + sin x | +C
= ⋅ ln
2
cos x
cos 2 x
2
1
sin x
−
+C
cos x cox
= ln | sec x + tan x | + C
= ln
- 55 -
88.
x
+ C = ln csc x − cot x + C
2
x
x
x
sin 2 + cos 2
1 + tan 2
1
1
2
2 =
2
证法1：∵ csc x =
=
=
x
x
x
x
x
sin x
2 ⋅ sin ⋅ cos
2 ⋅ sin ⋅ cos
2 tan
2
2
2
2
2
x 1
1
又 ∵ d tan = ⋅
dx
2 2
2 x
cos
2
x
x
∴ dx = 2 ⋅ cos 2 d tan
2
2
1
x
x
∴ ∫ csc x dx = ∫
⋅2 ⋅ cos 2 d tan
x
x
2
2
2 ⋅ sin ⋅ cos
2
2
1
x
=∫
d tan
x
2
tan
2
x
= ln tan + C
2
x
x
x
sin
sin 2
2 sin 2
x
1 − cos x
2 =
2
2
∵ tan =
=
=
= csc x − cot x
x
x
x
x
x
2
sin x
cos
sin ⋅ cos
2 sin ⋅ cos
2
2
2
2
2
x
∴ ∫ csc x dx = ln tan + C = ln csc x − cot x + C
2
1
证法2 ：∫ csc x dx = ∫
dt
sint
sint
=∫
dt
sin 2 t
1
= −∫
d cos t
1 − cos 2 t
1
1
1
= − ∫(
+
)d cos t
2 1 + cost 1 − cost
1
1
1
1
=− ∫
d (cos t + 1) + ∫
d (1 − cos t )
2 1 + cost
2 1 − cost
1
1
= − ⋅ ln 1 + cost + ⋅ ln cost − 1 + C1
2
2
1
cost − 1
= ⋅ ln
+ C1
2
1 + cost
∫ csc x dx = ln
tan
1
(1 − cos t ) 2
= ⋅ ln
⋅ ( −1) + C1
2
1 − cos 2 t
=
1
(1 − cos t ) 2
⋅ ln
+ C2
2
sin 2 t
= ln
1 − cost
+ C2
sint
= ln csc x − cot x + C
- 56 -
89.
∫ sec
2
x dx = tan x + C
证明：∵ (tan x ) ′ = sec 2 x 即 tan x为 sec 2 x的原函数
∴ ∫ sec 2 x dx = ∫ d tan t
= tan x + C
90.
∫ csc
2
x dx = −cotx + C
证明： ∫ csc 2 x dx = − ∫ ( −csc 2 x ) dx
∵ (cotx) ′ = − csc 2 x 即 cot x为 − csc 2 x的原函数
∴ ∫ csc 2 x dx = − ∫ d cotx
= −cotx + C
91.
∫ sec x ⋅ tan x dx = sec x + C
证明：∵ ( sec x) ′ = sec x ⋅ tan x 即 sec x为 sec x ⋅ tan x的原函数
∴ ∫ sec x ⋅ tan x dx = ∫ d sec x
= sec x + C
92.
∫ cscx ⋅ cot x dx = − csc x + C
证明： ∫ cscx ⋅ cot x dx = − ∫ ( −cscx ⋅ cot x ) dx
∵ (csc x ) ′ = −cscx ⋅ cot x 即 csc x为 − cscx ⋅ cot x的原函数
∴ ∫ cscx ⋅ cot x dx = − ∫ d csc x
= − csc x + C
93.
94.
x 1
− ⋅ sin 2 x + C
2 4
1 1
证明：∫ sin 2 x dx = ∫ ( − ⋅ cos2x) dx
2 2
1
1
= ∫ dx − ∫ cos2x d 2 x
2
4
x 1
= − sin 2 x + C
2 4
∫ sin
2
x dx =
x 1
+ ⋅ sin 2 x + C
2 4
1 1
证明：∫ cos 2 x dx = ∫ ( + ⋅ cos2x) dx
2 2
1
1
= ∫ dx + ∫ cos2x d 2 x
2
4
x 1
= + sin 2 x + C
2 4
∫ cos
2
提示：sin 2 x =
1 − cos 2 x
2
提示：
cos 2 x =
1 + cos 2 x
2
x dx =
- 57 -
95.
∫ sin
n
1
n −1
x dx = − ⋅ sin n −1 x ⋅ cos x +
sin n − 2 x dx
∫
n
n
证明：∫ sin n x dx = ∫ sin n −1 x ⋅ sin x dx
= − ∫ sin n −1 x d cos x
= − cos x ⋅ sin n −1 x + ∫ cos x d sin n −1 x
= − cos x ⋅ sin n −1 x + ∫ cos x ⋅ ( n − 1) ⋅ sin n− 2 x ⋅ cos x dx
= − cos x ⋅ sin n −1 x + ( n − 1) ∫ cos 2 x ⋅ sin n −2 x dx
= − cos x ⋅ sin n −1 x + ( n − 1) ∫ (1 − sin 2 x ) ⋅ sin n − 2 x dx
= − cos x ⋅ sin n −1 x + ( n − 1) ∫ sin n − 2 x dx − ( n − 1) ∫ sin n x dx
移项并整理得：n ∫ sin n x dx = − cos x ⋅ sin n −1 x + (n − 1) ∫ sin n − 2 x dx
∴
96.
∫ cos
n
x dx =
∫ sin
n
1
n −1
x dx = − ⋅ sin n −1 x ⋅ cos x +
sin n − 2 x dx
∫
n
n
1
n −1
⋅ cos n −1 x ⋅ sin x +
cos n − 2 x dx
∫
n
n
证明：∫ cos n x dx = ∫ cos n −1 x ⋅ cos x dx
= ∫ cos n −1 x d sin x
= sin x ⋅ cos n−1 x − ∫ sin x d cos n−1 x
= sin x ⋅ cos n−1 x + ∫ sin x ⋅ ( n − 1) ⋅ cos n− 2 x ⋅ sin x dx
= sin x ⋅ cos n−1 x + ( n − 1) ∫ sin 2 x ⋅s cos n − 2 x dx
= sin x ⋅ cos n−1 x + ( n − 1) ∫ (1 − cos 2 x ) ⋅ cos n − 2 x dx
= sin x ⋅ cos n−1 x + ( n − 1) ∫ cos n − 2 x dx − ( n − 1) ∫ cos n x dx
移项并整理得：n ∫ cos n x dx = sin x ⋅ cos n −1 x + ( n − 1) ∫ cos n −2 x dx
∴
- 58 -
∫ sin
n
x dx =
1
n −1
⋅ sin x ⋅ cos n −1 x +
cos n −2 x dx
∫
n
n
dx
1
cos x
n−2
dx
⋅
+
n −1
∫
n − 1 sin x n − 1 sin n − 2 x
x
dx
1
1
证明：∫
dx = − ∫
⋅
dx
n
n−2
sin x
sin x − sin 2 x
1
= −∫
d cot x
sin n −2 x
cot x
1
=−
+ ∫ cot x d
n−2
sin x
sin n − 2 x
cot x
=−
+ cot x ⋅ (2 − n) ⋅ sin 1− n x ⋅ cos x dx
sin n − 2 x ∫
cot x
cos 2 x
=−
+ (2 − n) ∫
dx
sin n − 2 x
sin n x
cot x
1 − sin 2 x
=−
+
(2
−
n
)
∫ sin n x dx
sin n − 2 x
cot x
dx
1
=−
+ (2 − n) ∫
dx − (2 − n) ∫
dx
n−2
n
sin x
sin x
sin n − 2 x
dx
cot x
1
移项并整理得：
( n − 1) ∫
dx = − n − 2 − (2 − n) ∫
dx
n
sin x
sin x
sin n −2 x
cos x
1
=−
+ ( n − 2) ∫
dx
n −1
sin x
sin n −2 x
dx
1
cos x
n−2
dx
∴ ∫
dx = −
⋅
+
n
n −1
∫
n − 1 sin x n − 1 sin n − 2 x
sin x
97.
∫ sin
98.
∫ cos
n
dx
dx = −
1
sin x
n−2
dx
⋅
+
n −1
∫
n − 1 cos x n − 1 cos n − 2 x
x
dx
1
1
证明：∫
=∫
⋅
dx
n
n−2
cos x
cos x cos 2 x
1
=∫
d tan x
cos n − 2 x
tan x
1
=
+ ∫ tan x d
n−2
cos x
cos n −2 x
tan x
=
+ ∫ tan x ⋅ (2 − n) ⋅ cos1− n x ⋅ sin x dx
n−2
cos x
tan x
sin 2 x
=
− ( n − 2) ∫
dx
cos n − 2 x
cos n x
tan x
1 − cos 2 x
=
−
(
n
−
2
)
∫ cos n x dx
cos n − 2 x
sin x
dx
1
=
− ( n − 2) ∫
dx + (n − 2)∫
dx
n −1
n
cos x
cos x
cos n − 2 x
dx
sin x
1
移项并整理得：
( n − 1) ∫
=
+ ( n − 2) ∫
dx
n
n −1
cos x cos x
cos n − 2 x
sin x
1
=
+ ( n − 2) ∫
dx
n −1
cos x
cos n − 2 x
dx
1
sin x
n−2
dx
∴ ∫
=−
⋅
+
n
n −1
∫
n − 1 cos x n − 1 cos n − 2 x
cos x
n
=−
- 59 -
99.
1
m −1
⋅ cos m−1 x ⋅ sin n+1 x +
cos m−2 x ⋅ sin n xdx
①
m+n
m+n∫
1
n −1
=−
⋅ cos m+1 x ⋅ sin n−1 x +
cos m x ⋅ sin n−2 xdx
②
∫
m+n
m+n
证明①：∵ d sin m+ n xdx = (m + n) ⋅ sin m +n−1 x ⋅ cos xdx
1
∴ ∫ cos m x ⋅ sin n xdx =
cos m −1 x ⋅ sin1−m xd sin m+ n x
∫
m+n
1
1
=
⋅ cos m−1 x ⋅ sin n+1 x −
sin m+n xd (cos m−1 x ⋅ sin1−m x)
∫
m+ n
m+n
m−1
1− m
m− 2
1− m
∵ d (cos x ⋅ sin x) = [−(m − 1) ⋅ cos x ⋅ sin x ⋅ sin x + (1 − m) ⋅ sin1−m−1 x ⋅ cos x ⋅ cos m−1 x]dx
∫ cos
m
x ⋅ sin n xdx =
= [(1 − m) ⋅ sin −m x ⋅ cos m x ⋅ ( sin 2 x ⋅ cos −2 x + 1)]dx
sin 2 x + cos 2 x
)]dx
cos 2 x
= [(1 − m) ⋅ sin −m x ⋅ cos m −2 x]dx
1
m −1
∴ −
sin m +n xd (cos m−1 x ⋅ sin1−m x) =
cos m−2 x ⋅ sin n xdx
∫
∫
m+n
m+n
1
m −1
∴ ∫ cos m x ⋅ sin n xdx =
⋅ cos m−1 x ⋅ sin n+1 x +
cos m−2 x ⋅ sin n xdx
∫
m+n
m+ n
= [(1 − m) ⋅ sin −m x ⋅ cos m x ⋅ (
证明②：∵ d cos m+n x = −(m + n) ⋅ cos m+n−1 x ⋅ sin xdx
−1
∴ ∫ cos m x ⋅ sin n xdx =
cos1−n x ⋅ sin n−1 xd cos m +n x
∫
m+n
−1
1
=
⋅ sin n−1 x ⋅ cos m+1 x +
cos m+n xd (sin n−1 x ⋅ cos1−n x)
∫
m+ n
m+n
n −1
1− n
n−2
1− n
∵ d (sin x ⋅ cos x) = [(n − 1) ⋅ sin x ⋅ cos x ⋅ cos x − (1 − n) ⋅ cos1−n−1 x ⋅ sin x ⋅ sin n−1 x]dx
= [(n− 1) ⋅ cos −n x⋅ sin n x⋅ ( sin −2 x⋅ cos 2 x + 1)]dx
sin 2 x + cos 2 x
)]dx
sin 2 x
= [(n − 1) ⋅ cos − n x ⋅ sin n−2 x]dx
1
n −1
∴
cos m+n xd (sin n−1 x ⋅ cos1−n x) =
cos m x ⋅ sin n−2 xdx
∫
∫
m+n
m+n
1
n −1
∴ ∫ cos m x ⋅ sin n xdx = −
⋅ cos m+1 x ⋅ sin n−1 x +
cos m x ⋅ sin n−2 xdx
∫
m+n
m+ n
= [(n − 1) ⋅ cos − n x ⋅ sin n x ⋅ (
- 60 -
100.
1
1
∫ sin ax ⋅ cos bx dx = − 2(a + b) ⋅ cos(a + b) x − 2(a − b) ⋅ cos(a − b) x + C
1
1
证明：∫ sin ax ⋅ cos bx dx = ∫ [ sin ( a + b) x + sin ( a − b) x ]dx 提示： sin α cos β = 2 [ sin ( α + β ) + sin
2
1
1
= ∫ sin ( a + b) x dx ∫ sin ( a − b) xdx
2
2
1
1
=
sin (a + b) x d (a + b ) x +
sin (a − b) x d (a − b ) x
∫
2(a + b)
2( a − b) ∫
=−
101.
1
( α − β )]
1
1
⋅ cos( a + b ) x −
⋅ cos( a − b) x
2( a + b)
2( a − b )
1
∫ sin ax ⋅ sin bx dx = − 2(a + b) ⋅ sin (a + b ) x + 2(a − b) ⋅ sin (a − b) x + C
1
1
证明：∫ sin ax ⋅ sin bx dx = ∫ [cos ( a − b) x − cos( a + b) x]dx 提示：sin α sin β = − 2 [ cos ( α + β ) + cos
2
1
1
= ∫ cos ( a − b) x dx − ∫ cos ( a + b) x dx
2
2
1
1
=
cos (a − b) x d (a − b ) x −
cos (a + b) x d (a + b) x
∫
2(a − b)
2(a + b) ∫
=
102.
1
( α − β )]
1
1
⋅ sin ( a − b ) x −
⋅ sin ( a + b) x + C
2( a − b)
2( a + b)
1
∫ cos ax ⋅ cos bx dx = 2(a + b) ⋅ sin (a + b) x + 2(a − b) ⋅ sin (a − b) x + C
1
1
证明：∫ cos ax ⋅ cos bx dx = ∫ [cos ( a + b) x + cos ( a − b) x ]dx 提示：cos α cos β = [ cos ( α + β ) + cos ( α − β )]
2
2
1
1
= ∫ cos ( a + b) x dx + ∫ cos ( a − b) x dx
2
2
1
1
=
cos (a + b) x d (a + b) x +
cos (a − b) x d (a − b) x
∫
2(a + b)
2( a − b) ∫
=
1
1
⋅ sin ( a + b) x +
⋅ sin (a − b ) x + C
2( a + b)
2( a − b)
- 61 -
103.
dx
∫ a + b ⋅ sin x =
x
a ⋅ tan + b
2
2
⋅ arctan
+C
2
2
2
a −b
a − b2
(a 2 > b 2 )
x
2 = 2t
x 1+ t 2
1 + tan 2
2
x
1
x
1
x
1
dt = (tan )dx = ⋅ sec 2 dx = (1 + tan 2 ) dx = (1 + t 2 ) dx
2
2
2
2
2
2
2
2bt
a (1 + t 2 ) + 2bt
∴ dx =
dt
,
a
+
b
⋅
sin
x
=
a
+
=
1+ t 2
1+ t 2
1+ t2
dx
1+ t2
2
∴∫
=∫
⋅
dt
2
a + b ⋅ sin x
a (1 + t ) + 2bt 1 + t 2
1
= 2∫ 2
dt
a t + 2bt + a
1
= 2∫
dt
b 2 b2
a (t + ) −
+a
a
a
1
= 2a ∫
dt
2
( at + b) + ( a 2 − b 2 )
1
= 2∫
d (at + b)
2
( at + b) + ( a 2 − b 2 )
2 ⋅ tan
x
x
x
证明：
令 t = tan , 则 sin x = 2 ⋅ sin ⋅ cos =
2
2
2
当a 2 > b 2 ，
即a 2 − b 2 > 0 时
2∫
1
1
d (at + b) = 2∫
d (at + b)
2
2
2
( at + b) + ( a − b )
( at + b) + ( a 2 − b 2 ) 2
公式 19 ：∫
2
dx
1
x
= ⋅ arctan + C
x + a2 a
a
2
2
=
a −b
x
dx
将t = tan 代入上式得：
=
∫
2
a + b sin x
- 62 -
2
2
⋅ arctan
2
a2 − b2
at + b
a2 − b2
x
+b
2
+C
a2 − b2
a ⋅ tan
⋅ arctan
+C
x
a ⋅ tan + b − b 2 − a 2
dx
1
2
104. ∫
=
⋅ ln
+C
(a 2 < b 2 )
2
2
x
a + b sin x
b −a
a ⋅ tan + b + b 2 − a 2
2
x
2 ⋅ tan
x
x
x
2 = 2t
证明：
令t = tan , 则 sin x = 2 ⋅ sin ⋅ cos =
x 1+ t 2
2
2
2
1 + tan 2
2
x
1
x
1
x
1
dt = (tan )dx = ⋅ sec 2 dx = (1 + tan 2 ) dx = (1 + t 2 ) dx
2
2
2
2
2
2
2
2
2bt
a (1 + t ) + 2bt
∴ dx =
dt , a + b sin x = a +
=
2
2
1+ t
1+ t
1+ t2
dx
1+ t2
2
∴∫
=∫
⋅
dt
2
a + b sin x
a (1 + t ) + 2bt 1 + t 2
1
= 2∫ 2
dt
a t + 2bt + a
1
= 2∫
dt
b 2 b2
a (t + ) −
+a
a
a
1
= 2a ∫
dt
2
( at + b) + ( a 2 − b 2 )
1
= 2∫
d (at + b)
2
( at + b) + ( a 2 − b 2 )
当a 2 < b 2 ，
即a 2 − b 2 < 0 时
2∫
1
1
d (at + b ) = 2 ∫
d (at + b)
2
2
2
( at + b ) + ( a − b )
( at + b) − (b 2 − a 2 )
1
= 2∫
d (at + b)
( at + b) 2 − ( b 2 − a 2 ) 2
公式 21 ：∫
2
dx
x2 − a
2
=
1
x − a
⋅ ln
+ C
2a
x + a
= 2×
1
2 b2 − a2
⋅ ln
at + b − b 2 − a 2
at + b + b 2 − a 2
+C
x
a ⋅ tan + b − b 2 − a 2
x
dx
1
2
将t = tan 代入上式得：
=
⋅ ln
+C
∫
2
2
x
2
a + b sin x
2
2
b −a
a ⋅ tan + b + b − a
2
- 63 -
105.
dx
2
∫ a + b ⋅ cos x = a + b ⋅
⎛ a−b
a+b
x⎞
arctan ⎜⎜
⋅ tan ⎟⎟ + C
a−b
2⎠
⎝ a+b
(a 2 > b 2 )
x
2
x
2 = 1− t
证明：令 t = tan ，
则 cos x =
x 1+ t2
2
1 + tan 2
2
2
1− t
( a + b ) + t 2 ( a − b)
∴ a + b ⋅ cos x = a + b ⋅
=
1+ t2
1+ t 2
x 1
1
1
1+ t 2
2 x
∵ dt = d tan = ⋅ sec dx =
dx =
dx =
dx
2 2
2
1 + cos x
2
2 x
2 cos
2
2
1 + cos 2 θ
∴ dx =
dt
提示 ：cos 2 θ =
2
1+ t
2
dx
2
∴ ∫
=∫
dt
a + b ⋅ cos x
( a + b ) + t 2 ( a − b)
1 − tan 2
当 | a | >| b | ，即 a 2 > b 2 时
2
∫ ( a + b) + t
2
( a − b)
dt =
2
1
dt
∫
a − b ⎛ a + b ⎞2
2
⎜
⎟
⎜ a−b ⎟ +t
⎝
⎠
公式 19 ：∫
=
2
a−b
⋅
⋅ arctan
a−b a+b
⎛ a−b ⎞
⎜
⎟
⎜ a + b ⋅t⎟ + C
⎝
⎠
=
2
a−b
⋅
⋅ arctan
a−b a+b
⎛ a−b ⎞
⎜
⎟
⎜ a + b ⋅t⎟ + C
⎝
⎠
=2
=
1
⋅ arctan
(a + b ) ⋅ (a − b )
2
a+b
⋅
⋅ arctan
a+b a−b
dx
1
x
= ⋅ arctan + C
a
x2 + a2 a
⎛ a−b ⎞
⎜
⎟
⎜ a + b ⋅t⎟ +C
⎝
⎠
⎛ a−b ⎞
⎜
⎟
⎜ a + b ⋅t⎟ + C
⎝
⎠
⎛ a−b
x
dx
2
a+b
x⎞
将 t = tan 代入上式得：
=
⋅
arctan ⎜⎜
⋅ tan ⎟⎟ + C
∫
2
a + b ⋅ cos x a + b a − b
2⎠
⎝ a+b
- 64 -
x
a+b
+
dx
1
a+b
2
b−a
106. ∫
=
⋅
⋅ ln
+C
(a 2 < b 2 )
a + b ⋅ cos x a + b b − a
x
a+b
tan −
2
b−a
x
1 − tan 2
2
x
2 = 1− t
证明：令t = tan ，
则 cos x =
x 1+ t 2
2
1 + tan 2
2
2
1− t
( a + b) + t 2 ( a − b)
∴ a + b ⋅ cos x = a + b ⋅
=
1+ t 2
1+ t2
x 1
x
1
1
1+ t 2
∵ dt = d tan = ⋅ sec 2 dx =
dx =
dx =
dx
2 2
2
1 + cos x
2
2 x
2 cos
2
2
∴ dx =
dt
1 + cos 2 θ
提示 ：cos 2 θ =
1+ t 2
2
dx
2
∴ ∫
=
dt
a + b ⋅ cos x ∫ (a + b) + t 2 (a − b)
tan
当 a 2 < b 2 , 即 | a | <| b | ，∴ b − a > 0
2
2
dt = ∫
dt
( a − b)
( a + b ) − t 2 (b − a )
2
1
2
1
=
dt =
dt
∫
2
∫
2
b−a ⎛ a+b ⎞
a−b
⎛
⎞
a+b
2
⎜
⎟
⎟
t 2 − ⎜⎜
⎜ b−a ⎟ −t
⎟
b
−
a
⎝
⎠
⎝
⎠
∫ (a + b ) + t
2
a+b
a+b
t−
2 1 b−a
1
b−a
b−a
b−a
=
⋅ ⋅
⋅ ln
+C =
⋅
⋅ ln
+C
a−b 2 a+b
a−b a+b
a+b
a+b
t+
t+
b−a
b−a
1
x −a
t−
公式 21 ：∫
dx
=
⋅ ln
+ C
2a
x +a
x2 − a2
a+b
a+b
t−
1
1
a+b
b−a
b−a
= ( −1)
⋅ ln
+C =−
⋅
⋅ ln
+C
(a + b ) ⋅ (b − a )
a+b b−a
a+b
a+b
t+
t+
b−a
b−a
t−
a+b
1
a+b
b−a
=
⋅
⋅ ln
+C
a+b b−a
a+b
t−
b−a
t+
x
tan +
x
dx
1
a+b
2
将t = tan 代入上式得：
=
⋅
⋅ ln
∫
2
a + b ⋅ cos x a + b b − a
x
tan −
2
a+b
b−a +C
a+b
b−a
- 65 -
107.
108.
dx
1
⎛b
⎞
=
⋅ arctan ⎜ ⋅ tan x ⎟ + C
2
2
cos x + b sin x ab
⎝a
⎠
dx
1
1
证明：∫ 2
=∫
⋅ 2
dx
2
2
2
2
a cos x + b sin x
cos x a + b 2 tan 2 x
1
=∫ 2
d tan x
a + b 2 tan 2 x
1
1
= 2∫ 2
d tan x
b
a
2
( 2 + tan x)
b
1
1
= 2∫
d tan x
a 2
b
2
( ) + tan x )
b
1 b
⎛b
⎞
= 2 ⋅ ⋅ arctan ⎜ ⋅ tan x ⎟ + C
b a
⎝a
⎠
1
⎛b
⎞
=
⋅ arctan ⎜ ⋅ tan x ⎟ + C
ab
⎝a
⎠
∫a
2
2
dx
1
x
= ⋅ arctan + C
2
a
a
x +a
2
dx
1
b ⋅ tan x + a
=
⋅ ln
+C
2
2
b ⋅ tan x − a
cos x − b sin x 2ab
dx
1
1
证明：∫ 2
=∫
⋅ 2
dx
2
2
2
2
a cos x − b sin x
cos x a − b 2 tan 2 x
1
=∫ 2
d tan x
a − b 2 tan 2 x
1
1
= ∫ 2
d (b ⋅ tan x)
b a − (b ⋅ tan x ) 2
1
1
=− ∫
d (b ⋅ tan x)
b (b ⋅ tan x) 2 − a 2
∫a
2
2
1 1
b ⋅ tan x − a
=− ⋅
⋅ ln
+C
b 2a
b ⋅ tan x + a
=−
=
- 66 -
公式 19 ：∫
1
b ⋅ tan x − a
⋅ ln
+C
2ab
b ⋅ tan x + a
1
b ⋅ tan x + a
⋅ ln
+C
2ab
b ⋅ tan x − a
公式 21 ：∫
dx
1
x −a
=
⋅ ln
+ C
2a
x +a
x2 − a2
提示：log a b −1 = − log a b
109.
1
∫ x ⋅ sin ax dx = a
2
⋅ sin ax −
1
⋅ x ⋅ cos ax + C
a
1
x d cos ax
a∫
1
1
= − ⋅ x ⋅ cos ax + ∫ cos ax dx
a
a
1
1
= − ⋅ x ⋅ cos ax + 2 ∫ cos ax dax
a
a
1
1
= − ⋅ x ⋅ cos ax + 2 ⋅ sin ax + C
a
a
证明：∫ x ⋅ sin ax dx = −
1 2
2
2
⋅ x ⋅ cos ax + 2 ⋅ x ⋅ sin ax + 3 ⋅ cos ax + C
a
a
a
1
证明：∫ x 2 ⋅ sin ax dx = − ∫ x 2 d cos ax
a
1
1
= − ⋅ x 2 ⋅ cos ax + ∫ cos ax dx 2
a
a
1
2
= − ⋅ x 2 ⋅ cos ax + ∫ x ⋅ cos ax dx
a
a
1
2
= − ⋅ x 2 ⋅ cos ax + 2 ⋅ ∫ x d sin ax
a
a
1
2
2
= − ⋅ x 2 ⋅ cos ax + 2 ⋅ x ⋅ sin ax − 3 ⋅ ∫ sin ax dax
a
a
a
1
2
2
= − ⋅ x 2 ⋅ cos ax + 2 ⋅ x ⋅ sin ax + 3 ⋅ cos ax
a
a
a
2
110.
∫x
⋅ sin ax dx = −
111.
∫ x ⋅ cos ax dx = a
1
2
⋅ cos ax −
1
⋅ x ⋅ sin ax + C
a
1
x d sin ax
a∫
1
1
= ⋅ x ⋅ sin ax − ∫ sin ax dx
a
a
1
1
= ⋅ x ⋅ sin ax − 2 ∫ sin ax dax
a
a
1
1
= ⋅ x ⋅ sin ax + 2 ⋅ cos ax + C
a
a
证明：∫ x ⋅ cos ax dx =
112.
1
2
2
⋅ cos ax dx = ⋅ x 2 ⋅ sin ax + 2 ⋅ x ⋅ cos ax − 3 ⋅ sin ax + C
a
a
a
1
证明：∫ x 2 ⋅ cos ax dx = ∫ x 2 d sin ax
a
1
1
= ⋅ x 2 ⋅ sin ax − ∫ sin ax dx 2
a
a
1
2
= ⋅ x 2 ⋅ sin ax + ∫ x ⋅ sin ax dx
a
a
1
2
= ⋅ x 2 ⋅ sin ax − 2 ⋅ ∫ x d cos ax
a
a
1
2
2
= ⋅ x 2 ⋅ sin ax + 2 ⋅ x ⋅ cos ax − 3 ⋅ ∫ cos ax dax
a
a
a
1
2
2
= ⋅ x 2 ⋅ sin ax + 2 ⋅ x ⋅ cos ax − 3 ⋅ sin ax + C
a
a
a
∫x
2
- 67 -
（十二）含有反三角函数的积分（其中 a > 0)（113~121）
113.
x
x
∫ arcsin a dx = x ⋅ arcsin a +
证明：∫ arcsin
a2 − x2 + C
( a > 0)
x
x
x
dx = x ⋅ arcsin − ∫ x d arcsin
a
a
a
x
1
1
= x ⋅ arcsin − ∫ x ⋅
⋅ dx
a
a
x
1− ( )2
a
x
x
= x ⋅ arcsin − ∫
dx
2
a
a − x2
= x ⋅ arcsin
x 1
1
− ∫
dx 2
2
2
a 2
a −x
1
−
x 1
+ ∫ (a 2 − x 2 ) 2 d (a 2 − x 2 )
a 2
1
1−
x 1 1
2
2
= x ⋅ arcsin + ⋅
⋅ (a − x ) 2 + C
1
a 2
1−
2
x
= x ⋅ arcsin + a 2 − x 2 + C
a
2
2
x
x
a
x x 2
114. ∫ x ⋅ arcsin dx = ( − ) ⋅ arcsin +
a − x2 + C
( a > 0)
a
2
4
a 4
x
证明：
令 t = arcsin , 则 x = a ⋅ sin t
a
x
∴ ∫ x ⋅ arcsin dx = ∫ a ⋅ sin t ⋅ t d ( a ⋅ sin t ) = a 2 ∫ t ⋅ sin t ⋅ cost dt
a
a2
a2
=
t
⋅
sin
2
t
dt
=
−
t d cos 2t
2 ∫
4 ∫
a2
a2
=−
⋅ t ⋅ cos 2t +
cos 2t dt
4
4 ∫
a2
a2
=−
⋅ t ⋅ cos 2t +
cos 2t d 2t
4
8 ∫
a2
a2
=−
⋅ t ⋅ cos 2t +
⋅ sin 2t + C
4
8
提示：sin 2 x = 2 ⋅ sin x ⋅ cos x
2
2
a
a
cos 2 x = cos 2 x − sin 2 x
=−
⋅ t ⋅ (2 cos 2 t − 1) +
⋅ sin t ⋅ cos t + C
4
4
= 2 cos 2 x − 1
2
2
2
a
a
a
=−
⋅ t ⋅ cos 2 t +
⋅t +
⋅ sin t ⋅ cos t + C
2
4
4
= x ⋅ arcsin
在Rt ΔABC中，可设 ∠B = t ,| AB |= a ，
则 | AC |= x ,| BC |= a 2 − x 2
∴ cos t =
∴ ∫ x ⋅ arcsin
- 68 -
a2 − x2
x
, sin t =
a
a
x
a2
x a2 − x2 a2
x a2 x a2 − x2
dx = − ⋅ arcsin ⋅
+
⋅
arcsin
+
⋅ ⋅
+C
a
2
a
a2
4
a 4 a
a
x2 − a2
x a2
x x
=
⋅ arcsin +
⋅ arcsin + ⋅ a 2 − x 2 + C
2
a 4
a 4
2
2
x
a
x x 2
= ( − ) ⋅ arcsin +
a − x2 + C
2
4
a 4
x
x3
x 1
115. ∫ x ⋅ arcsin dx =
⋅ arcsin + ( x 2 + 2a 2 ) a 2 − x 2 + C
( a > 0)
a
3
a 9
x
证明：
令t = arcsin , 则 x = a ⋅ sin t
a
x
∴ ∫ x 2 ⋅ arcsin dx = ∫ a 2 ⋅ sin 2 t ⋅ t d ( a ⋅ sin t ) = a 3 ∫ t ⋅ sin 2 t ⋅ cost dt
a
a3
=
t d sin 3 t
∫
3
a3
a3
3
=
⋅ t ⋅ sin t −
sin 3 t dt
∫
3
3
3
a
a3
=
⋅ t ⋅ sin 3 t −
sin t (1 − cos 2 t ) dt
∫
3
3
3
a
a3
a3
3
=
⋅ t ⋅ sin t −
sin t dt +
sin t ⋅ cos 2 t dt
∫
∫
3
3
3
3
3
3
a
a
a
=
⋅ t ⋅ sin 3 t +
⋅ cos t −
cos 2 t d cos t
3
3
3 ∫
a3
a3
a3 1
3
=
⋅ t ⋅ sin t +
⋅ cos t −
⋅
⋅ cos 3 t + C
3
3
3 1+ 2
3
3
a
a
a3
=
⋅ t ⋅ sin 3 t +
⋅ cos t −
⋅ cos 3 t + C
3
3
9
2
在Rt ΔABC中，可设 ∠B = t ,| AB |= a ，
则 | AC |= x ,| BC |= a 2 − x 2
∴ cos t =
a2 − x2
x
, sin t =
a
a
x
a3
x x3 a3
a2 − x2 a3 a 2 − x2
∴ ∫ x ⋅ arcsin dx =
⋅ arcsin ⋅ 3 +
⋅
−
⋅
⋅ a 2 − x2 + C
3
a
3
a a
3
a
9
a
3
2
2
2
x
x a
a −x
=
⋅ arcsin +
⋅ a2 − x2 −
⋅ a2 − x2 + C
3
a 3
9
3
x
x 1
=
⋅ arcsin + ( x 2 + 2a 2 ) a 2 − x 2 + C
3
a 9
2
- 69 -
116.
x
x
∫ arccos a dx = x ⋅ arccos a −
证明：∫ arccos
a2 − x2 + C
( a > 0)
x
x
x
dx = x ⋅ arccos − ∫ x d arccos
a
a
a
x
1
1
= x ⋅ arccos + ∫ x ⋅
⋅ dx
a
a
x
1− ( )2
a
x
x
= x ⋅ arccos + ∫
dx
2
a
a − x2
= x ⋅ arccos
x 1
1
+ ∫
dx 2
2
2
a 2
a −x
1
−
x 1
− ∫ (a 2 − x 2 ) 2 d (a 2 − x 2 )
a 2
1
1−
x 1 1
= x ⋅ arccos − ⋅
⋅ (a 2 − x 2 ) 2 + C
1
a 2
1−
2
x
= x ⋅ arccos − a 2 − x 2 + C
a
= x ⋅ arccos
117 .
x
x2 a2
x x
dx = (
−
) ⋅ arccos −
a2 − x2 + C
( a > 0)
a
2
4
a 4
x
证明：令 t = arccos
, 则 x = a ⋅ cos t
a
x
∴ ∫ x ⋅ arccos dx = ∫ a ⋅ cos t ⋅ t d ( a ⋅ cos t ) = − a 2 ∫ t ⋅ cos t ⋅ sin t dt
a
a2
a2
=−
t ⋅ sin 2 t dt =
t d cos 2t
2 ∫
4 ∫
a2
a2
=
⋅ t ⋅ cos 2t −
cos 2t dt
4
4 ∫
a2
a2
提示：
sin 2 x = 2 ⋅ sin x ⋅ cos x
=
⋅ t ⋅ cos 2t −
cos 2t d 2t
∫
4
8
cos 2 x = cos 2 x − sin 2 x
2
2
a
a
=
⋅ t ⋅ cos 2t −
⋅ sin 2t + C
= 2 cos 2 x − 1
4
8
a2
a2
=
⋅ t ⋅ (2 cos 2 t − 1) −
⋅ sin t ⋅ cos t + C
4
4
a2
a2
a2
2
=
⋅ t ⋅ cos t −
⋅t −
⋅ sin t ⋅ cos t + C
2
4
4
∫ x ⋅ arccos
在 Rt Δ ABC 中，可设 ∠ B = t ,| AB |= a ，
则 | BC |= x ,| AC |=
∴ sin t =
∴
- 70 -
∫ x ⋅ arccos
a2 − x2
a2 − x2
x
, cos t =
a
a
x
a2
x x2 a2
x a2 x
a2 − x2
dx =
⋅ arcsin ⋅ 2 −
⋅ arcsin −
⋅ ⋅
+C
a
2
a a
4
a
4 a
a
x2
x a2
x x
=
⋅ arcsin −
⋅ arcsin − ⋅ a 2 − x 2 + C
2
a
4
a 4
2
2
x
a
x x
=(
−
) ⋅ arcsin +
a2 − x2 + C
2
4
a 4
118.
x
x3
x 1
2
x
⋅
arccos
dx
=
⋅ arccos − ( x 2 + 2 a 2 ) a 2 − x 2 + C
( a > 0)
∫
a
3
a 9
x
证明：
令 t = arccos , 则 x = a ⋅ cos t
a
x
∴ ∫ x 2 ⋅ arccos dx = ∫ a 2 ⋅ cos 2 t ⋅ t d ( a ⋅ cos t ) = −a 3 ∫ t ⋅ cos 2 t ⋅ sin t dt
a
a3
=
t d cos 3 t
∫
3
a3
a3
=
⋅ t ⋅ cos 3 t −
cos 3 t dt
3
3 ∫
a3
a3
=
⋅ t ⋅ cos 3 t −
cos t (1 − sin 2 t ) dt
∫
3
3
3
a
a3
a3
=
⋅ t ⋅ cos 3 t −
cos
t
dt
+
cos t ⋅ sin 2 t dt
∫
∫
3
3
3
3
3
3
a
a
a
=
⋅ t ⋅ cos 3 t −
⋅ sin t +
sin 2 t d sin t
∫
3
3
3
3
3
a
a
a3 1
3
=
⋅ t ⋅ cos t −
⋅ sin t +
⋅
⋅ sin 3 t + C
3
3
3 1+ 2
3
3
a
a
a3
=
⋅ t ⋅ cos 3 t −
⋅ sin t +
⋅ sin 3 t + C
3
3
9
在Rt ΔABC中，可设 ∠B = t ,| AB |= a ，
则 | BC |= x ,| AC |= a 2 − x 2
∴ sin t =
∴ ∫ x 2 ⋅ arccos
119.
x
a2 − x2
x
, cos t =
a
a
x
a3
x x3 a 3
a 2 − x 2 a3 a 2 − x 2
dx =
⋅ arcsin ⋅ 3 −
⋅
+
⋅
⋅ a 2 − x2 + C
a
3
a a
3
a
9
a3
x3
x a2
a2 − x2
=
⋅ arcsin −
⋅ a2 − x2 +
⋅ a2 − x2 + C
3
a 3
9
3
x
x 1
=
⋅ arcsin − ( x 2 + 2a 2 ) a 2 − x 2 + C
3
a 9
x
a
∫ arctan a dx = x ⋅ arctan a − 2 ⋅ ln (a
2
+ x2 ) + C
( a > 0)
x
x
x
dx = x ⋅ arctan − ∫ x d x ⋅ arctan
a
a
a
x
1
1
= x ⋅ arctan − ∫ x ⋅
⋅ dx
x
a
a
1 + ( )2
a
x
x
= x ⋅ arctan − a ∫ 2
dx
a
a + x2
x a
1
= x ⋅ arctan − ∫ 2
dx 2
a 2 a + x2
x a
1
= x ⋅ arctan − ∫ 2
d (a 2 + x 2 )
a 2 a + x2
x a
= x ⋅ arctan − ⋅ ln a 2 + x 2 + C
a 2
2
2
∵ a +x >0
x
x a
∴ ∫ arctan dx = x ⋅ arctan − ⋅ ln ( a 2 + x 2 ) + C
a
a 2
证明：∫ arctan
- 71 -
120.
x
1
∫ x ⋅ arctan a dx = 2 (a
2
+ x 2 ) ⋅ arctan
x a
− ⋅x+C
a 2
( a > 0)
x
, 则 x = a ⋅ tan t
a
x
∴ ∫ x ⋅ arctan dx = ∫ a ⋅ tan t ⋅ t d ( a ⋅ tan t ) = a 2 ∫ t ⋅ sec 2 t ⋅ tan t dt
a
a2
=
t d sec 2 t
∫
2
a2
a2
2
=
⋅ t ⋅ sec t −
sec 2 t dt
∫
2
2
2
a
a2
=
⋅ t ⋅ sec 2 t −
⋅ tan t + C
2
2
证明：
令t = arctan
在Rt ΔABC中，可设 ∠B = t ,| BC |= a ，
则 | AC |= x ,| AB |= a 2 + x 2
1
a2 + x2
x
∴ sec t =
=
, tan t =
cos t
a
a
2
2
2
x
a
x a +x
a2 x
∴ ∫ x ⋅ arctan dx =
⋅ arctan ⋅
−
⋅ +C
a
2
a
a2
2 a
1
x a
= ( a 2 + x 2 ) ⋅ arctan − ⋅ x + C
2
a 2
121.
x
x3
x a 2 a3
2
x
⋅
arctan
dx
=
⋅
arctan
− ⋅x +
ln (a 2 + x 2 ) + C
( a > 0)
∫
a
3
a 6
6
x
1
x
证明：∴ ∫ x 2 ⋅ arctan dx = ∫ arctan dx 3
a
3
a
3
x
x 1
1
1
=
⋅ arctan − ∫ x 3 ⋅
⋅ dx
x
3
a 3
a
1+ ( )2
a
3
3
x
x a
x
=
⋅ arctan − ∫ 2
dx
3
a 3 a + x2
x3
x a
x2
=
⋅ arctan − ∫ 2
dx 2
2
3
a 6 a +x
3
x
x a x2 + a2 − a2 2
=
⋅ arctan − ∫
dx
3
a 6
a2 + x2
x3
x a
a
a2
2
=
⋅ arctan − ∫ dx + ∫ 2
dx 2
2
3
a 6
6 a +x
3
x
x a
a3
1
=
⋅ arctan − ∫ dx 2 +
d(x2 + a2 )
2
2
∫
3
a 6
6 a +x
3
3
x
x a
a
=
⋅ arctan − ⋅ x 2 +
ln a 2 + x 2 + C
3
a 6
6
2
2
∵ a +x >0
∴ ∫ x 2 ⋅ arctan
- 72 -
x
x3
x a
a3
dx =
⋅ arctan − ⋅ x 2 + ln ( a 2 + x 2 ) + C
a
3
a 6
6
（十三）含有指数函数的积分（122~131）
122.
1
⋅ax + C
ln a
1
证明：∫ a x dx =
ln a ⋅ a x dx
∫
ln a
x
x
∵ ( a ) ′ = a ln a , 即 a x ln a的原函数为a x
∫a
x
dx =
1
da x
∫
ln a
1
=
⋅ ax + C
ln a
∴ ∫ a x dx =
123.
∫e
ax
dx =
1 ax
⋅e + C
a
μ
1
, dx = dμ
a
a
1
1
∴ ∫ e ax dx = ∫ e μ dμ = ⋅ e μ + C
a
a
1
= ⋅ e ax + C
a
证明：
令 ax = μ , 则 x =
124.
125.
1
( ax − 1) e ax + C
2
a
1
证明：∫ x ⋅ e ax dx = ∫ x de ax
a
1
1
= ⋅ x ⋅ e ax − ∫ e ax dx
a
a
1
1
= ⋅ x ⋅ e ax − 2 ∫ e ax dax
a
a
1
1
= ⋅ x ⋅ e ax − 2 e ax + C
a
a
1
= 2 ( ax − 1)e ax + C
a
∫ x ⋅e
ax
dx =
1 n ax n n −1 ax
⋅ x ⋅ e − ∫ x ⋅ e dx
a
a
1
证明：∫ x n ⋅ e ax dx = ∫ x n de ax
a
1
1
= ⋅ x n ⋅ e ax − ∫ e ax dx n
a
a
1
n
= ⋅ x n ⋅ e ax − ∫ x n −1 ⋅ e ax dx
a
a
∫x
n
⋅ e ax dx =
- 73 -
126.
127.
128.
- 74 -
x
1
⋅ax −
⋅ax + C
ln a
(ln a) 2
1
证明：∫ x ⋅ a x dx =
x da x
∫
ln a
1
1
1
x
公式122：∫ a x dx =
⋅ax + C
=
⋅ x ⋅ ax −
a
dx
∫
ln a
ln a
ln a
1
1
=
⋅ x ⋅ ax −
⋅ ax + C
2
ln a
(ln a)
∫ x⋅a
x
dx =
1
n
⋅ xn ⋅ a x −
x n −1 ⋅ a x dx
ln a
ln a ∫
1
证明：∫ x n ⋅ a x dx =
x n da x
ln a ∫
1
1
=
⋅ xn ⋅ a x −
a x dx n
∫
ln a
ln a
1
n
=
⋅ xn ⋅ a x −
x n−1 ⋅ a x dx
∫
ln a
ln a
∫x
n
⋅ a x dx =
1
⋅ e ax ( a ⋅ sin bx − b ⋅ cos bx) + C
2
a +b
1
证明：∫ e ax ⋅ sin bx dx = − ∫ e ax d cos bx
b
1
1
= − ⋅ e ax ⋅ cos bx + ∫ cos bxde ax
b
b
1
a
a
= − ⋅ e ax ⋅ cos bx + 2 ⋅ e ax ⋅ sin bx − 2 ∫ sin bx de ax
b
b
b
1
a
a
= − ⋅ e ax ⋅ cos bx + 2 ⋅ e ax ⋅ sin bx − 2 ∫ sin bx de ax
b
b
b
2
2
a +b
1
a
移项并整理得： 2 ∫ e ax ⋅ sin bx dx = − ⋅ e ax ⋅ cos bx + 2 ⋅ e ax ⋅ sin bx + C
b
b
b
b
a
∴ ∫ e ax ⋅ sin bx dx = − 2
⋅ e ax ⋅ cos bx + 2
⋅ e ax ⋅ sin bx + C
2
2
a +b
a +b
1
= 2
⋅ e ax ( a ⋅ sin bx − b ⋅ cos bx ) + C
a + b2
∫e
ax
⋅ sin bx dx =
2
129.
1
⋅ e ax (b ⋅ sin bx + a ⋅ cos bx ) + C
2
a +b
1
证明：∫ e ax ⋅ cos bxdx = ∫ e ax d sin bx
b
1
1
= ⋅ e ax ⋅ sin bx − ∫ sin bxde ax
b
b
1
a
= ⋅ e ax ⋅ sin bx − ∫ sin bx ⋅ e ax dx
b
b
1
a
= ⋅ e ax ⋅ sin bx + 2 ∫ e ax d cos bx
b
b
1
a
a
= ⋅ e ax ⋅ sin bx + 2 ⋅ e ax ⋅ cos bx − 2 ∫ cos bxde ax
b
b
b
1
a
a2
= ⋅ e ax ⋅ sin bx + 2 ⋅ e ax ⋅ cos bx − 2 ∫ e ax ⋅ cos bxdx
b
b
b
2
2
2
a
a +b
1
a
∴ (1 + 2 ) ∫ e ax ⋅ cos bxdx =
e ax ⋅ cos bxdx = ⋅ e ax ⋅ sin bx + 2 ⋅ e ax ⋅ cos bx
2
∫
b
b
b
b
1
∴ ∫ e ax ⋅ cos bxdx = 2
⋅ e ax (b ⋅ sin bx + a ⋅ cos bx) + C
a + b2
∫e
ax
⋅ cos bxdx =
2
- 75 -
130.
∫e
ax
⋅ sin n bx dx =
1
⋅e ax ⋅ sin n −1 bx( a ⋅ sin bx − nb ⋅ cos bx )
2 2
a +b n
n ⋅ (n − 1)b 2
+ 2
e ax ⋅ sin n − 2 bx dx
a + b2 n2 ∫
2
证明：∫ e ax ⋅ sin n bx dx = ∫ e ax ⋅ sin n − 2 bx ⋅ sin 2 bx dx = ∫ e ax ⋅ sin n − 2 bx ⋅ (1 − cos 2 bx ) dx
= ∫ e ax ⋅ sin n − 2 bx dx − ∫ e ax ⋅ sin n − 2 bx ⋅ cos 2 bx dx
又
∫e
ax
⋅ sin n − 2 bx ⋅ cos 2 bx dx =
=
①
1
e ax ⋅ cos bx d sin n −1 bx
∫
b ⋅ (n − 1)
1
1
⋅ e ax ⋅ cos bx ⋅ sin n −1 bx −
sin n −1 bx d (e ax ⋅ cos bx)
b ⋅ (n − 1)
b ⋅ (n − 1) ∫
②
又 ∫ sin n −1 bx d ( e ax ⋅ cos bx) = ∫ sin n −1 bx ( a ⋅ e ax ⋅ cos bx − b ⋅ sin bx ⋅ e ax ) dx
= a ∫ e ax ⋅ sin n −1 bx ⋅ cos bx dx − b ∫ sin n bx ⋅ e ax dx
③
1 ax
e ⋅ sin n −1 bx d sin bx
∫
b
1
1
= ⋅ e ax ⋅ sin n bx − ∫ sin bx d (e ax ⋅ sin n −1 bx)
b
b
1
1
= ⋅ e ax ⋅ sin n bx − ∫ sin bx [ a ⋅ e ax ⋅ sin n −1 bx + b ⋅ ( n − 1) sin n − 2 bx ⋅ cos bx ⋅ e ax ]dx
b
b
1
a
= ⋅ e ax ⋅ sin n bx − ∫ sin n bx ⋅ e ax dx − ( n − 1) ∫ e ax ⋅ sin n −1 bx ⋅ cos bx dx
b
b
1 ax
a
ax
n −1
n
n
ax
移项并整理得：
④
∫ e ⋅ sin bx ⋅ cos bx dx = bn ⋅ e ⋅ sin bx − bn ∫ sin bx ⋅ e dx
又
∫e
ax
⋅ sin n −1 bx ⋅ cos bx dx =
n −1
ax
将④式代入③式的得：
∫ sin bx d (e ⋅ cos bx)
a ax
a2
⋅ e ⋅ sin n bx −
sin n bx ⋅ e ax dx − b ∫ sin n bx ⋅ e ax dx
∫
bn
bn
a ax
a2 + b2n
=
⋅ e ⋅ sin n bx −
sin n bx ⋅ e ax dx
∫
bn
bn
1
ax
n−2
2
ax
n −1
将⑤式代入②式得：
∫ e ⋅ sin bx ⋅ cos bx dx = b ⋅ (n − 1) ⋅ e ⋅ cos bx ⋅ sin bx
=
⑤
a
a 2 + b2 n
ax
n
⋅
e
⋅
sin
bx
+
sin n bx ⋅ e ax dx
b 2 ⋅ n ⋅ (n − 1)
b 2 ⋅ n ⋅ (n − 1) ∫
1
ax
n
ax
n−2
ax
n −1
将 式代入①式得：
∫ e ⋅ sin bx dx = ∫ e ⋅ sin bx dx − b ⋅ (n − 1) ⋅ e ⋅ cos bx ⋅ sin bx
−
+
a
a2 + b2n
ax
n
⋅
e
⋅
sin
bx
−
sin n bx ⋅ e ax dx
2
2
∫
b ⋅ n ⋅ (n − 1)
b ⋅ n ⋅ (n − 1)
ax
n
移项并整理得：
∫ e ⋅ sin bx dx
⎤
n ⋅ (n − 1)b 2 ⎡ ax
1
1
e ⋅ sin n − 2 bx dx −
⋅ e ax ⋅ cos bx ⋅ sin n −1 bx +
⋅ e ax ⋅ sin n bx ⎥
2
2 2 ⎢∫
2
a +b n ⎣
b ⋅ (n − 1)
n ⋅ (n − 1)b
⎦
2
n ⋅ (n − 1)b
bn
a
= 2
⋅ ∫ e ax ⋅ sin n − 2 bx dx − 2
⋅ e ax ⋅ cos bx ⋅ sin n −1 bx + 2
⋅ e ax ⋅ sin n bx
2 2
2 2
2 2
a +b n
a +b n
a +b n
1
= 2
⋅ e ax ⋅ sin n −1 bx ( a ⋅ sin bx − nb ⋅ cos bx)
2 2
a +b n
n ⋅ (n − 1)b 2
+ 2
e ax ⋅ sin n −2 bx dx
2 2
∫
a +b n
=
- 76 -
131.
∫e
ax
⋅ cos n bx dx =
1
⋅ e ax ⋅ cos n −1 bx( a ⋅ cos bx + nb ⋅ sin bx)
2 2
a +b n
n ⋅ (n − 1)b 2
+ 2
e ax ⋅ cos n − 2 bx dx
a + b 2n2 ∫
2
证明：∫ e ax ⋅ cos n bx dx =∫ e ax ⋅ cos n− 2 bx ⋅ cos 2 bx dx =∫ e ax ⋅ cos n − 2 bx ⋅ (1 − sin 2 bx) dx
= ∫ e ax ⋅ cos n− 2 bx dx − ∫ e ax ⋅ cos n −2 bx ⋅ sin 2 bx dx
又
∫e
1
e ax ⋅ sin bx d cos n −1 bx
∫
b ⋅ (1 − n)
1
1
=
⋅ e ax ⋅ sin bx ⋅ cos n −1 bx −
cos n −1 bx d (e ax ⋅ sin bx)
b ⋅ (1 − n)
b ⋅ (1 − n) ∫
ax
①
⋅ cos n− 2 bx ⋅ sin 2 bx dx =
②
又 ∫ cos n −1 bx d (e ax ⋅ sin bx) = ∫ cos n −1 bx ( a ⋅ e ax ⋅ sin bx + b ⋅ cos bx ⋅ e ax ) dx
= a ∫ e ax ⋅ cos n−1 bx ⋅ sin bx dx + b ∫ cos n bx ⋅ e ax dx
又
∫e
ax
⋅ cos n −1 bx ⋅ sin bx dx = −
③
1 ax
e ⋅ cos n −1 bx d cos bx
∫
b
1
1
= − ⋅ e ax ⋅ cos n bx + ∫ cos bx d (e ax ⋅ cos n −1 bx)
b
b
1
1
= − ⋅ e ax ⋅ cos n bx + ∫ cos bx [ a ⋅ e ax ⋅ cos n −1 bx − b ⋅ ( n − 1) cos n − 2 bx ⋅ sin bx ⋅ e ax ]dx
b
b
1
a
= − ⋅ e ax ⋅ cos n bx + ∫ cos n bx ⋅ e ax dx − ( n − 1) ∫ e ax ⋅ cos n −1 bx ⋅ sin bx dx
b
b
1 ax
a
ax
n −1
n
n
ax
移项并整理得：
④
∫ e ⋅ cos bx ⋅ sin bx dx = − bn ⋅ e ⋅ cos bx + bn ∫ cos bx ⋅ e dx
n −1
ax
将④式代入③式的得：
∫ cos bx d (e ⋅ sin bx)
a ax
a2
⋅ e ⋅ cos n bx +
cos n bx ⋅ e ax dx + b ∫ cos n bx ⋅ e ax dx
∫
bn
bn
a
a 2 + b2n
= − ⋅ e ax ⋅ cos n bx +
cos n bx ⋅ e ax dx
∫
bn
bn
1
ax
n−2
2
ax
n −1
将⑤式代入②式得：
∫ e ⋅ cos bx ⋅ sin bx dx = b ⋅ (1 − n) ⋅ e ⋅ sin bx ⋅ cos bx
=−
⑤
a
a2 + b2n
ax
n
⋅
e
⋅
cos
bx
−
cos n bx ⋅ e ax dx
b 2 ⋅ n ⋅ (1 − n)
b 2 ⋅ n ⋅ (1 − n) ∫
1
ax
n
ax
n−2
ax
n −1
将 式代入①式得：
∫ e ⋅ cos bx dx = ∫ e ⋅ cos bx dx − b ⋅ (1 − n) ⋅ e ⋅ sin bx ⋅ cos bx
+
+
a
a 2 + b2n
ax
n
⋅
e
⋅
cos
bx
−
cos n bx ⋅ e ax dx
2
2
∫
b ⋅ n ⋅ (n − 1)
b ⋅ n ⋅ (n − 1)
ax
n
移项并整理得：
∫ e ⋅ cos bx dx
⎤
n ⋅ (1 − n)b 2 ⎡ ax
1
a
e ⋅ cos n −2 bx dx −
⋅ e ax ⋅ sin bx ⋅ cos n −1 bx −
⋅ e ax ⋅ cos n bx⎥
2
2 2 ⎢∫
2
b ⋅ (1 − n )
−a −b n ⎣
n ⋅ (1 − n)b
⎦
2
n ⋅ (n − 1)b
bn
a
= 2
⋅ ∫ e ax ⋅ cos n − 2 bx dx + 2
⋅ e ax ⋅ sin bx ⋅ cos n −1 bx + 2
⋅ e ax ⋅ cos n bx
2 2
2 2
2 2
a +b n
a +b n
a +b n
2
1
n ⋅ (n − 1)b
= 2
⋅ e ax ⋅ cos n −1 bx( a ⋅ cos bx + nb ⋅ sin bx) + 2
e ax ⋅ cos n− 2 bx dx
2 2
a +b n
a + b 2n2 ∫
=
- 77 -
（十四）含有对数函数的积分（132~136）
132.
∫ ln xdx = x ⋅ ln x − x + C
证明：∫ ln xdx = x ⋅ ln x − ∫ x d ln x
= x ⋅ ln x − ∫ x ⋅
1
dx
x
= x ⋅ ln x − ∫ dx
= x ⋅ ln x − x + C
133.
dx
∫ x ⋅ ln x dx = ln
证明：∫
134.
- 78 -
ln x + C
dx
1
dx = ∫
d ln x
x ⋅ ln x
ln x
= ln ln x + C
提示：
(ln x) ′ =
1
1
⋅ x n +1 (ln x −
)+C
n +1
n +1
ln x
证明：∫ x n ⋅ ln x dx = ∫
⋅ ( n + 1) ⋅ x n dx
n +1
ln x
=∫
dx n +!
n +1
ln x n +1
1
=
⋅x −
x n +1 d ln x
∫
n +1
n +1
ln x n +1
1
=
⋅x −
x n dx
∫
n +1
n +1
ln x n +1
1 2 n +1
=
⋅x −(
) ⋅x +C
n +1
n +1
1
1
=
⋅ x n +1 (ln x −
)+C
n +1
n +1
∫x
n
⋅ ln x dx =
1
x
n
135.
∫ (lnx) dx = x ⋅ (lnx)
n
− n ∫ (lnx) n−1dx
n−k
n
= x∑ (−1)
⋅
k =0
n！
⋅ (lnx) k
k！
n
证明：∫ (lnx) dx = x ⋅ (lnx) n − ∫ xd( ln x)n
1
= x ⋅ (lnx) n − ∫ x ⋅ n ⋅ ( ln x)n−1 ⋅ dx
x
= x ⋅ (lnx) n − n∫ (lnx) n−1dx
= x ⋅ (lnx) n − n ⋅ x ⋅ (lnx) n−1 + n ∫ xd( ln x)n−1
= x ⋅ (lnx) n − n ⋅ x ⋅ (lnx) n−1 + n ⋅ (n − 1)∫ (lnx) n−2 dx
= x ⋅ (lnx) n − n ⋅ x ⋅ (lnx) n−1 + n ⋅ (n − 1) ⋅ x ⋅ (lnx) n −2 − n ⋅ (n − 1) ⋅ (n − 2)∫ (lnx) n−3 dx
.......
= x ⋅ (lnx) n − n ⋅ x ⋅ (lnx) n−1 + n ⋅ (n − 1) ⋅ x ⋅ (lnx) n −2 − n ⋅ (n − 1) ⋅ (n − 2)(lnx) n−3
+ ⋯⋯ + ( −1 )n−k ⋅ n ⋅ (n − 1) ⋅ (n − 2)⋯⋯( n − k + 1 ) ⋅ (lnx) n−k + ⋯⋯
+ (−1) 2 ⋅ n ⋅ (n − 1) ⋅ (n − 2) ⋯⋯5 × 4 × 3 ⋅ (lnx) 3−1 ⋅ x
+ (−1)1 ⋅ n ⋅ (n − 1) ⋅ (n − 2)⋯⋯4 × 3 × 2 ⋅ (lnx) 2−1 ⋅ x
+ (−1) 0 ⋅ n ⋅ (n − 1) ⋅ (n − 2)⋯⋯3 × 2 ×1 ⋅ (lnx)1−1 ⋅ x
n
= x∑ (−1)
k =0
136.
n−k
⋅
n！
⋅ (lnx) k
k！
1
n
⋅ x m +1 ⋅ (ln x) n −
x m ⋅ (ln x ) n −1 dx
m +1
m +1 ∫
1
证明：∫ x m ⋅ (ln x ) n dx =
(ln x ) n dx m +1
m +1 ∫
1
1
=
⋅ x m +1 ⋅ (ln x) n −
x m +1 d (ln x ) n
∫
m +1
m +1
1
n
1
=
⋅ x m +1 ⋅ (ln x) n −
x m +1 ⋅ (ln x ) n −1 ⋅ dx
∫
m +1
m +1
x
1
n
=
⋅ x m +1 ⋅ (ln x) n −
x m ⋅ (ln x ) n −1 dx
∫
m +1
m +1
∫x
m
⋅ (ln x ) n dx =
- 79 -
（十五）含有双曲函数的积分（137~141）
137.
∫ shx dx = chx + C
证明：∵ (chx) ′ = shx ,即 chx为 shx的原函数
∴ ∫ shx dx = ∫ d chx
= chx + C
138.
∫ ch x dx = shx + C
证明：∵ ( shx ) ′ = chx ,即 shx为chx的原函数
∴ ∫ ch x dx = ∫ d shx
= shx + C
139.
∫ th x dx = ln chx + C
shx
dx
chx
1
=∫
d chx
chx
= ln chx + C
证明：∫ th x dx = ∫
140.
∫ sh
2
x dx = −
x 1
+ sh 2 x + C
2 4
2
e x + e−x
⎛ e x − e −x ⎞
提示：chx =
⎟⎟ dx
证明：∫ sh x dx = ∫ ⎜⎜
2
2 ⎠
⎝
x
e − e−x
1
2x
−2 x
shx
=
= ∫ ( e + e − 2) dx
2
4
2x
−2 x
e
e
x
=
−
− +C
8
8
2
2x
x 1 e − e −2 x
=− + ⋅
+C
2 4
2
x 1
= − + ⋅ sh 2 x + C
2 4
2
141.
∫ ch
2
x dx =
2
- 80 -
(双曲余弦)
x 1
+ ⋅ sh 2 x + C
2 4
⎛ e x + e −x ⎞
⎟⎟ dx
证明：∫ ch x dx = ∫ ⎜⎜
2
⎝
⎠
1
= ∫ ( e 2 x + e − 2 x + 2) dx
4
e 2 x e −2 x x
=
−
+ +C
8
8
2
2x
x 1 e − e −2 x
= + ⋅
+C
2 4
2
x 1
= + ⋅ sh 2 x + C
2 4
2
(双曲余弦)
e x + e −x
2
x
e − e −x
shx =
2
提示：chx =
(双曲余弦)
(双曲余弦)
（十六）定积分 （142~147）
142.
∫
π
π
cos nx dx = ∫ sin nx dx = 0
−π
−π
1 π
cos nx dnx
n ∫−π
1
π
= ⋅ ( sin nx −π )
n
1
1
= ⋅ sin ( nπ ) − ⋅ sin ( −nπ )
n
n
2
= ⋅ sin ( nπ )
n
=0
π
证明①：∫ cos nx dx =
−π
1 π
sin nx dnx
n ∫− π
1
π
= − ⋅ ( cos nx − π )
n
1
1
= − ⋅ cos ( nπ ) + ⋅ cos ( − nπ )
n
n
=0
π
证明②：∫ sin nx dx =
−π
π
π
−π
−π
综合证明①②得：∫ cos nx dx = ∫ sin nx dx = 0
143.
∫
π
cos mx ⋅ sin nx dx = 0
−π
证明：1.当 m ≠ n时
公式 100 ：∫ sin ax ⋅ cos bx dx = −
1
1
⋅ cos ( a + b ) x −
⋅ cos ( a − b ) x + C
2 (a + b )
2(a − b )
π
π
1
1
∫−πcos mx ⋅ sin nx dx = − 2(m + n) ⋅ cos(m + n ) x −π − 2(n − m) cos(n − m) x −π
π
1
1
[cos( m + n)π − cos( m + n) π ] −
[cos( n − m) π − cos ( n − m)( −π )]
2( m + n)
2( n − m )
=0+0 = 0
=−
2.当 m = n时
∫
π
π
cos mx ⋅ sin nx dx = ∫ cos mx ⋅ sin mx dx
−π
−π
提示：
sin 2 x = 2 ⋅ sin x ⋅ cos x
1 π
sin 2mx d mx
2 m ∫−π
1 π
=
sin 2mx d 2mx
4 m ∫−π
=
π
1
=−
⋅ cos 2mx
4m
−π
=−
1
⋅ [ cos 2mπ − cos( −2 mπ )]
4m
=0
π
π
−π
−π
综合讨论 1 , 2 得：∫ cos nx dx = ∫ cos mx ⋅ sin nx dx = 0
- 81 -
144.
⎧0 ，m ≠ n
cos mx ⋅ cos nx dx = ⎨
−π
⎩π，m = n
证明：1.当 m ≠ n时
∫
π
π
π
1
1
∫− πcos mx ⋅ cos nx dx = 2(m + n) ⋅ sin (m + n) x − 2(m − n) sin ( m − n) x
−π
−π
π
1
1
[ sin ( m + n) π − sin ( m + n )( −π )] −
[ sin ( m − n )π + sin ( m − n)( −π )]
2( m + n )
2( m − n )
=
= 0−0 = 0
公式 102 ：∫ cos ax ⋅ cos bx dx =
2.当 m = n时
∫
π
1
1
⋅ sin (a + b ) x +
⋅ sin ( a − b ) x + C
2 (a + b )
2( a − b )
π
cos mx ⋅ cos nx dx = ∫ cos mx ⋅ cos mx dx
−π
−π
=
1 π
cos 2 mx d mx
∫
−
π
m
=
1
1
⋅ sin 2 mx +
⋅ mx
4m
2m
−π
−π
公式 94 ：∫ cos
π
2
x dx =
x 1
+ ⋅ sin 2 x + C
2 4
π
1
π π
⋅ [ sin 2m π − sin ( −2mπ )] + +
4m
2 2
=π
=
π
⎧0 ，m ≠ n
综合讨论 1 , 2 得：∫ cos mx ⋅ cos nx dx = ⎨
−π
⎩π，m = n
145.
⎧0 ，m ≠ n
sin mx ⋅ sin nx dx = ⎨
−π
⎩π，m = n
证明：1.当 m ≠ n时
∫
π
π
π
1
1
∫−πsin mx ⋅ sin nx dx = − 2(m + n) ⋅ sin (m + n) x + 2(m − n) sin (m − n ) x
−π
−π
π
=−
1
1
[ sin ( m + n )π − sin ( m + n)( −π )] +
[ sin ( m − n )π − sin ( m − n )( −π )]
2( m + n)
2( m − n)
= 0+0 = 0
2.当 m = n时
∫
π
−π
公式 101 ：∫ sin ax ⋅ sin bx dx = −
1
1
⋅ sin ( a + b ) x +
⋅ sin ( a − b ) x + C
2( a + b )
2( a − b )
π
sin mx ⋅ sin nx dx = ∫ sin 2 mx dx
−π
公式 93 ：∫ sin 2 x dx =
=
1 π 2
sin mx d mx
m ∫− π
=
1
1
⋅ mx −
⋅ sin 2 mx
2m
4m
−π
−π
π
=−
1
π π
⋅ [ sin 2mπ − sin ( −2 mπ )] + +
4m
2 2
=π
π
⎧0 ，m ≠ n
综合讨论 1 , 2 得：∫ sin mx ⋅ sin nx dx = ⎨
−π
⎩π，m = n
- 82 -
π
x
1
− ⋅ sin 2 x + C
2
4
146.
∫
π
0
sin mx ⋅ sin nx dx = ∫
π
0
⎧0 ，m ≠ n
⎪
cos mx ⋅ cos nx dx ⎨ π
⎪⎩ 2 ，m = n
证明：1.当m ≠ n时
π
∫
π
0
π
1
1
sin mx ⋅ sin nx dx = −
⋅ sin ( m + n) x +
sin (m − n) x
2( m + n)
2
(
m
−
n
)
0
0
1
1
[ sin ( m + n)π − sin 0] +
[ sin ( m − n )π − sin 0]
2( m + n)
2( m − n)
= 0+0 =0
=−
π
∫
π
0
π
1
1
cos mx ⋅ cos nx dx =
⋅ sin ( m + n) x +
sin (m − n ) x
2( m + n)
2( m − n)
0
0
=
1
1
[ sin ( m + n) π − sin 0] +
[ sin ( m − n )π + sin 0]
2( m + n)
2( m − n)
= 0+0 =0
2.当 m = n时
∫
π
0
π
sin mx ⋅ sin nx dx = ∫ sin 2 mx dx
0
=
1 π 2
sin mx d mx
m ∫0
=
1
1
⋅ mx −
⋅ sin 2 mx
2m
4m
0
0
π
π
1
π
⋅ [ sin 2mπ − sin 0] + + 0
4m
2
π
=
2
=−
∫
π
0
π
cos mx ⋅ cos nx dx = ∫ cos mx ⋅ cos mx dx
0
=
1 π
cos 2 mx d mx
m ∫0
π
π
1
1
=
⋅ sin 2mx +
⋅ mx
4m
2m
0
0
1
π
⋅ [ sin 2mπ − sin 0] + + 0
4m
2
π
=
2
=
π
π
0
0
综合讨论 1 , 2 得：∫ sin mx ⋅ sin nx dx = ∫
⎧0 ，m ≠ n
⎪
cos mx ⋅ cos nx dx ⎨ π
⎪⎩ 2 ，m = n
以上所用公式：
1
1
⋅ sin ( a + b ) x +
⋅ sin ( a − b ) x + C
2 (a + b)
2( a − b )
1
1
公式 102 ：∫ cos ax ⋅ cos bx dx =
⋅ sin ( a + b ) x +
⋅ sin ( a − b ) x + C
2(a + b )
2(a − b )
x 1
公式 93 ：∫ sin 2 x dx = − ⋅ sin 2 x + C
2 4
x 1
公式 94 ：∫ cos 2 x dx = + ⋅ sin 2 x + C
2 4
公式 101 ：∫ sin ax ⋅ sin bx dx = −
- 83 -
π
π
147. I n = ∫ 2 sin n x dx = ∫ 2 cos n x dx
0
0
n −1
I n−2
n
4 2
⎧n −1 n − 3
⎪⎪ n ⋅ n − 2 ⋅ ⋯ ⋅ 5 ⋅ 3 ( n为大于1的正奇数) , I 1 = 1
=⎨
⎪ n − 1 ⋅ n − 3 ⋅ ⋯ ⋅ 3 ⋅ 1 ⋅ π ( n为正偶数) , I = π
0
⎪⎩ n n − 2
4 2 2
2
In =
π
π
2
0
π
2
1
n − 1 2 n−2
sin n x dx = − ⋅ sin n−1 x ⋅ cos x +
sin x dx
n
n ∫0
0
证明①：I n = ∫
π
1
π
π
n − 1 2 n−2
= − ( sin n −1 ⋅ cos − sin n −1 0 ⋅ cos 0) +
sin x dx
n
2
2
n ∫0
π
n − 1 2 n−2
n −1
=
sin x dx =
I n−2
∫
n 0
n
当n为正奇数时
π
n −1 n − 3
4 2
⋅
⋅ ⋯ ⋅ ⋅ ⋅ ∫ 2 sinx dx
n n−2
5 3 0
π
n −1 n − 3
4 2
=
⋅
⋅ ⋯ ⋅ ⋅ ⋅ ( − cos x ) 02
n n−2
5 3
n −1 n − 3
4 2
=
⋅
⋅⋯ ⋅ ⋅ ⋅1
n n−2
5 3
In =
π
π
特别的，当n = 1时，I n = ∫ 2 sinx dx = ( − cos x) 02 = 1
0
当n为正偶数时
π
n −1 n − 3
3 1
In =
⋅
⋅ ⋯ ⋅ ⋅ ⋅ ∫ 2 sin 0 x dx
n n−2
4 2 0
π
n −1 n − 3
3 1
=
⋅
⋅ ⋯ ⋅ ⋅ ⋅ ( x ) 02
n n−2
4 2
n −1 n − 3
3 1 π
=
⋅
⋅⋯ ⋅ ⋅ ⋅
n n−2
4 2 2
π
2
0
π
特别的，当n = 0时，I n = ∫ sin 0 x dx = ( x) 02 =
π
证明②：I n = ∫ 2 cos n x dx ⋯⋯亦同理可证
0
- 84 -
π
2
附录：常数和基本初等函数导数公式
(C为常数)
1. ( C ) ′ = 0
2. ( x μ ) ′ = μ⋅ x μ −1 (x ≠ 0)
3. (sinx )′ = cosx
4. ( cosx) ′ = − sinx
5. (tanx)′ = sec 2 x
6. (cotx )′ = − csc 2 x
7. ( secx)′ = secx⋅ tanx
8. (cscx) ′ = − cscx⋅ cotx
9. ( a x ) ′ = a x ⋅ lna
( a 为常数)
10 . ( e x ) ′ = e x
11. (log a x ) ′ =
12. (lnx ) ′ =
1
x⋅ lna
( a > 0)
1
x
1
13. ( arcsinx) ′ =
14. ( arccosx)′ =
15. ( arctanx) ′ =
1− x2
1
− 1 − x2
1
1 + x2
16. ( arccotx) ′ = −
1
1+ x2
- 85 -