1
sin
2
x
+
cos
2
x
=
1
,
tan
2
x
+
1
=
sec
2
x
,
cot
2
x
+
1
=
csc
2
x
\sin^2x+\cos^2x=1, \tan^2x+1=\sec^2x,\cot^2x+1=\csc^2x
sin 2 x + cos 2 x = 1 , tan 2 x + 1 = sec 2 x , cot 2 x + 1 = csc 2 x
2
cosh
2
x
−
sinh
2
x
=
1
,
sinh
(
x
±
y
)
=
sinh
x
⋅
cosh
y
±
sinh
y
⋅
cosh
x
,
cosh
(
x
±
y
)
=
cosh
x
⋅
cosh
y
,
sinh
2
x
=
2
sinh
x
⋅
cosh
x
,
cosh
2
x
=
cosh
2
x
+
sinh
2
x
=
2
cosh
2
x
−
1
\begin{aligned}&\cosh^2x-\sinh^2x=1,\\ &\sinh(x\pm y)=\sinh x\cdot\cosh y \pm \sinh y\cdot\cosh x, \cosh(x\pm y) = \cosh x\cdot\cosh y,\\ &\sinh2x = 2\sinh x\cdot \cosh x, \cosh 2x = \cosh^2x+\sinh^2x = 2\cosh^2x - 1\end{aligned}
cosh 2 x − sinh 2 x = 1 , sinh ( x ± y ) = sinh x ⋅ cosh y ± sinh y ⋅ cosh x , cosh ( x ± y ) = cosh x ⋅ cosh y , sinh 2 x = 2 sinh x ⋅ cosh x , cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 x − 1
3 积化和差公式
sin
α
cos
β
=
1
2
[
sin
(
α
+
β
)
+
sin
(
α
−
β
)
]
,
cos
α
sin
β
=
1
2
[
sin
(
α
+
β
)
−
sin
(
α
−
β
)
]
,
cos
α
cos
β
=
1
2
[
c
o
s
(
α
+
β
)
+
c
o
s
(
α
−
β
)
]
,
sin
α
sin
β
=
−
1
2
[
cos
(
α
+
β
)
−
cos
(
α
−
β
)
]
\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)],\cos\alpha\sin\beta = \frac12[\sin(\alpha+\beta) - \sin(\alpha-\beta)],\\ \cos\alpha\cos\beta = \frac12[cos(\alpha+\beta) + cos(\alpha - \beta)], \sin\alpha\sin\beta=-\frac12[\cos(\alpha+\beta) - \cos(\alpha-\beta)]
sin α cos β = 2 1 [ sin ( α + β ) + sin ( α − β ) ] , cos α sin β = 2 1 [ sin ( α + β ) − sin ( α − β ) ] , cos α cos β = 2 1 [ c o s ( α + β ) + c o s ( α − β ) ] , sin α sin β = − 2 1 [ cos ( α + β ) − cos ( α − β ) ]
4 和差化积公式
sin
α
+
sin
β
=
2
sin
(
α
+
β
2
)
cos
(
α
−
β
2
)
,
sin
α
−
sin
β
=
2
sin
(
α
−
β
2
)
cos
(
α
+
β
2
)
,
cos
α
+
cos
β
=
2
cos
(
α
+
β
2
)
cos
(
α
−
β
2
)
,
cos
α
−
cos
β
=
−
2
sin
(
α
+
β
2
)
sin
(
α
−
β
2
)
\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right), \sin\alpha - \sin\beta = 2\sin\left(\frac{\alpha-\beta}{2}\right) \cos\left(\frac{\alpha+\beta}{2}\right), \\ \cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right), \cos\alpha - \cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)
sin α + sin β = 2 sin ( 2 α + β ) cos ( 2 α − β ) , sin α − sin β = 2 sin ( 2 α − β ) cos ( 2 α + β ) , cos α + cos β = 2 cos ( 2 α + β ) cos ( 2 α − β ) , cos α − cos β = − 2 sin ( 2 α + β ) sin ( 2 α − β )
5 三角代换
a
2
−
x
2
→
x
=
a
sin
t
a
cos
t
,
a
2
+
x
2
→
x
=
a
tan
t
a
sec
t
,
x
2
−
a
2
→
x
=
a
sec
t
a
tan
t
(
a
>
0
,
t
∈
(
−
π
2
,
π
2
)
)
\sqrt{a^2-x^2}\xrightarrow{x=a\sin t}a\cos t, \sqrt{a^2+x^2} \xrightarrow{x=a\tan t}a\sec t, \sqrt{x^2-a^2}\xrightarrow{x=a\sec t}a\tan t \left( a>0,t\in\left( -\frac{\pi}{2} , \frac{\pi}{2} \right) \right)
a 2 − x 2
x = a sin t
a cos t , a 2 + x 2
x = a tan t
a sec t , x 2 − a 2
x = a sec t
a tan t ( a > 0 , t ∈ ( − 2 π , 2 π ) )
6
∫
a
2
−
x
2
d
x
=
a
2
2
arcsin
x
a
+
1
2
x
a
2
−
x
2
+
C
(
a
>
0
)
\int\sqrt{a^2-x^2}\mathrm{d}x = \frac{a^2}{2}\arcsin\frac xa + \frac12x\sqrt{a^2-x^2} + C(a>0)
∫ a 2 − x 2
d x = 2 a 2 arcsin a x + 2 1 x a 2 − x 2
+ C ( a > 0 )
7
∫
x
2
+
a
2
d
x
=
x
2
x
2
+
a
2
+
a
2
2
ln
∥
x
+
x
2
+
a
2
∥
+
C
(
a
>
0
)
\int\sqrt{x^2+a^2}\mathrm{d}x = \frac x2\sqrt{x^2+a^2} + \frac{a^2}2\ln \left\| x+\sqrt{x^2+a^2} \right\|+C(a>0)
∫ x 2 + a 2
d x = 2 x x 2 + a 2
+ 2 a 2 ln ∥ ∥ ∥ x + x 2 + a 2
∥ ∥ ∥ + C ( a > 0 )
8
∫
1
x
2
+
a
2
d
x
=
ln
∥
x
+
x
2
+
a
2
∥
+
C
,
∫
1
x
2
−
a
2
d
x
=
ln
∥
x
+
x
2
+
a
2
∥
(
a
>
0
)
特
别
地
,
∫
1
x
2
+
1
d
x
=
(
x
+
x
2
+
1
)
=
a
r
s
i
n
h
x
,
∫
1
x
2
−
1
d
x
=
ln
(
x
+
x
2
−
1
)
=
a
r
c
o
s
h
x
\int\frac{1}{\sqrt{x^2+a^2}}\mathrm{d}x = \ln\left\| x+ \sqrt{x^2+a^2} \right\| +C, \int\frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x = \ln\left\| x + \sqrt{x^2+a^2}\right\|(a>0)\\ 特别地,\int\frac{1}{\sqrt{x^2+1}}\mathrm{d}x=\left(x+\sqrt{x^2+1}\right) = \mathrm{arsinh}\, x,\int\frac{1}{\sqrt{x^2-1}}\mathrm{d}x = \ln\left( x+\sqrt{x^2-1}\right) = \mathrm{arcosh}\, x
∫ x 2 + a 2
1 d x = ln ∥ ∥ ∥ x + x 2 + a 2
∥ ∥ ∥ + C , ∫ x 2 − a 2
1 d x = ln ∥ ∥ ∥ x + x 2 + a 2
∥ ∥ ∥ ( a > 0 ) 特 别 地 , ∫ x 2 + 1
1 d x = ( x + x 2 + 1
) = a r s i n h x , ∫ x 2 − 1
1 d x = ln ( x + x 2 − 1
) = a r c o s h x
9
∫
sec
x
d
x
=
∫
1
cos
x
d
x
=
∫
sec
2
x
+
sec
x
tan
x
sec
x
+
tan
x
d
x
=
∫
1
sec
x
+
tan
x
d
(
sec
x
+
tan
x
)
=
ln
∥
sec
x
+
tan
x
∥
+
C
\int\sec x\mathrm{d}x = \int\frac{1}{\cos x}\mathrm{d}x = \int\frac{\sec^2x+\sec x \tan x}{\sec x + \tan x}\mathrm{d}x = \int\frac{1}{\sec x + \tan x}\mathrm{d}(\sec x + \tan x) = \ln \left\| \sec x + \tan x \right\| + C
∫ sec x d x = ∫ cos x 1 d x = ∫ sec x + tan x sec 2 x + sec x tan x d x = ∫ sec x + tan x 1 d ( sec x + tan x ) = ln ∥ sec x + tan x ∥ + C
10
∫
csc
x
d
x
=
∫
1
sin
x
d
x
=
ln
∥
tan
x
2
∥
+
C
=
1
2
ln
∥
1
−
cos
x
1
+
cos
x
∥
+
C
\int\csc x\mathrm{d}x = \int\frac{1}{\sin x}\mathrm{d}x = \ln\left\| \tan\frac x2 \right\| + C = \frac 12 \ln \left\| \frac{1-\cos x}{1 + \cos x}\right\| + C
∫ csc x d x = ∫ sin x 1 d x = ln ∥ ∥ ∥ tan 2 x ∥ ∥ ∥ + C = 2 1 ln ∥ ∥ ∥ ∥ 1 + cos x 1 − cos x ∥ ∥ ∥ ∥ + C
11
∫
tan
x
d
x
=
∫
sin
x
cos
x
d
x
=
−
∫
1
cos
x
d
(
cos
x
)
=
−
ln
∥
cos
x
∥
+
C
\int\tan x\mathrm{d}x = \int\frac{\sin x}{\cos x}\mathrm{d}x = -\int\frac{1}{\cos x}\mathrm{d}(\cos x) = -\ln \left\| \cos x\right\| +C
∫ tan x d x = ∫ cos x sin x d x = − ∫ cos x 1 d ( cos x ) = − ln ∥ cos x ∥ + C
12
∫
1
tan
x
d
x
=
∫
cos
x
sin
x
d
x
=
∫
1
sin
x
d
(
sin
x
)
=
ln
∥
sin
x
∥
+
C
\int\frac{1}{\tan x}\mathrm{d}x = \int\frac{\cos x}{\sin x}\mathrm{d}x = \int\frac{1}{\sin x}\mathrm{d}(\sin x) = \ln\left\| \sin x \right\| + C
∫ tan x 1 d x = ∫ sin x cos x d x = ∫ sin x 1 d ( sin x ) = ln ∥ sin x ∥ + C
13
∫
arctan
x
d
x
=
x
arctan
x
−
ln
∥
1
+
x
2
∥
+
C
\int\arctan x\mathrm{d}x = x\arctan x - \ln \left\| 1+x^2 \right\| +C
∫ arctan x d x = x arctan x − ln ∥ ∥ 1 + x 2 ∥ ∥ + C
14Wallis 公式
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
{
n
−
1
n
⋅
n
−
3
n
−
2
⋅
⋯
⋅
1
2
⋅
π
2
,
n
为
偶
数
n
−
1
n
⋅
n
−
3
n
−
2
⋅
⋯
⋅
2
3
⋅
1
,
n
为
奇
数
=
{
(
n
−
1
)
!
!
n
!
!
⋅
π
2
,
n
为
偶
数
(
n
−
1
)
!
!
n
!
!
,
n
为
奇
数
\int^{\frac{\pi}{2}}_{0}\sin^nx\mathrm{d}x = \int^{\frac{\pi}{2}}_{0}\cos^nx\mathrm{d}x = \begin{cases} \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12\cdot\frac\pi2,&n为偶数 \\ \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23\cdot1,&n为奇数 \end{cases} = \begin{cases} \frac{(n-1)!!}{n!!}\cdot\frac\pi2,&n为偶数 \\ \frac{(n-1)!!}{n!!},&n为奇数 \end{cases}
∫ 0 2 π sin n x d x = ∫ 0 2 π cos n x d x = { n n − 1 ⋅ n − 2 n − 3 ⋅ ⋯ ⋅ 2 1 ⋅ 2 π , n n − 1 ⋅ n − 2 n − 3 ⋅ ⋯ ⋅ 3 2 ⋅ 1 , n 为 偶 数 n 为 奇 数 = { n ! ! ( n − 1 ) ! ! ⋅ 2 π , n ! ! ( n − 1 ) ! ! , n 为 偶 数 n 为 奇 数
15
∫
0
π
sin
n
x
d
x
=
2
∫
0
π
2
sin
n
x
d
x
,
∫
0
π
cos
n
x
d
x
=
{
2
∫
0
π
2
cos
n
x
d
x
,
n
为
偶
数
0
,
n
为
奇
数
∫
0
2
π
sin
n
x
=
∫
0
2
π
cos
n
x
d
x
=
{
4
∫
0
π
2
sin
n
x
d
x
,
n
为
偶
数
0
,
n
为
奇
数
∫
0
π
2
f
(
sin
x
)
d
x
=
∫
0
π
2
f
(
cos
x
)
d
x
,
∫
0
π
f
(
sin
x
)
d
x
≠
∫
0
π
f
(
cos
x
)
d
x
∫
0
π
π
f
(
sin
x
)
d
x
=
π
2
∫
0
π
f
(
sin
x
)
d
x
=
π
∫
0
π
2
f
(
sin
x
)
d
x
\int_0^\pi\sin^nx\mathrm{d}x = 2\int^\frac\pi2_0\sin^nx\mathrm{d}x,\int_0^\pi\cos^nx\mathrm{d}x = \begin{cases} 2\int^\frac\pi2_0\cos^nx\mathrm{d}x,&n为偶数 \\ 0,&n为奇数 \end{cases} \\ \int_0^{2\pi}\sin^nx = \int_0^{2\pi}\cos^nx\mathrm{d}x = \begin{cases} 4\int^\frac\pi2_0\sin^nx\mathrm{d}x,&n为偶数 \\ 0,&n为奇数 \end{cases} \\ \int_0^\frac\pi2f(\sin x)\mathrm{d}x = \int_0^\frac\pi2f(\cos x)\mathrm{d}x, \int_0^\pi f(\sin x)\mathrm{d}x \ne \int_0^\pi f(\cos x)\mathrm{d}x \\ \int_0^\pi\pi f(\sin x)\mathrm{d}x = \frac\pi2\int_0^\pi f(\sin x)\mathrm{d}x = \pi\int_0^\frac\pi2 f(\sin x)\mathrm{d}x
∫ 0 π sin n x d x = 2 ∫ 0 2 π sin n x d x , ∫ 0 π cos n x d x = { 2 ∫ 0 2 π cos n x d x , 0 , n 为 偶 数 n 为 奇 数 ∫ 0 2 π sin n x = ∫ 0 2 π cos n x d x = { 4 ∫ 0 2 π sin n x d x , 0 , n 为 偶 数 n 为 奇 数 ∫ 0 2 π f ( sin x ) d x = ∫ 0 2 π f ( cos x ) d x , ∫ 0 π f ( sin x ) d x = ∫ 0 π f ( cos x ) d x ∫ 0 π π f ( sin x ) d x = 2 π ∫ 0 π f ( sin x ) d x = π ∫ 0 2 π f ( sin x ) d x
16 万能公式
令
u
=
tan
x
2
,
x
=
2
arctan
u
,
则
sin
x
=
2
u
1
+
u
2
,
cos
x
=
1
−
u
2
1
+
u
2
,
d
x
=
2
1
+
u
2
d
u
令u=\tan\frac x2,x=2\arctan u,则\sin x = \frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}, \mathrm{d}x = \frac{2}{1+u^2}\mathrm{d}u
令 u = tan 2 x , x = 2 arctan u , 则 sin x = 1 + u 2 2 u , cos x = 1 + u 2 1 − u 2 , d x = 1 + u 2 2 d u
17
∫
a
cos
x
+
b
sin
x
c
cos
x
+
d
sin
x
=
A
∫
d
x
+
B
∫
d
(
c
cos
x
+
d
sin
x
)
c
cos
x
+
d
sin
x
(
设
a
cos
x
+
b
sin
x
=
A
(
c
cos
x
+
d
sin
x
)
+
B
(
c
cos
x
+
d
sin
x
)
′
,
由
待
定
系
数
法
求
出
A
、
B
)
\int \frac{a\cos x+ b\sin x}{c\cos x + d \sin x} = A\int\mathrm{d}x+B\int\frac{\mathrm{d}(c\cos x+d \sin x)}{c \cos x + d \sin x} \\ (设a\cos x + b\sin x = A(c \cos x + d \sin x) + B(c \cos x + d \sin x)',由待定系数法求出A、B)
∫ c cos x + d sin x a cos x + b sin x = A ∫ d x + B ∫ c cos x + d sin x d ( c cos x + d sin x ) ( 设 a cos x + b sin x = A ( c cos x + d sin x ) + B ( c cos x + d sin x ) ′ , 由 待 定 系 数 法 求 出 A 、 B )