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2024年3月21日发(作者:oracle minus)

_x0001_

微积分公式

D

x

sin x=cos x

cos x = -sin x

tan x = sec

2

x

cot x = -csc

2

x

sec x = sec x tan x

C

csc x = -csc x cot x

C

1

x

D

x

sin

-1

()=

22

a

ax

sin x dx = -cos x + C

cos x dx = sin x + C

tan x dx = ln |sec x | + C

cot x dx = ln |sin x | + C

sin

-1

(-x) = -sin

-1

x

cos

-1

(-x) = - cos

-1

x

tan

-1

(-x) = -tan

-1

x

cot

-1

(-x) = - cot

-1

x

- sec

-1

x sec x dx = ln |sec x + tan x | + sec

-1

(-x) =

csc

-1

(-x) = - csc

-1

x

csc x dx = ln |csc x – cot x | +

sin

-1

x dx = x sin

-1

x+

1x

2

+C

cos

-1

x dx = x cos

-1

x-

1x

2

+C

sinh

-1

(

x

)= ln (x+

a

2

x

2

) x

R

a

cos

-1

(

1

x

)=

22

a

ax

tan

-1

x dx = x tan

-1

x-½ln (1+x

2

)+C

x

-1

cosh ()=ln (x+

x

2

a

2

) x≧1

-1-12

cot x dx = x cot x+½ln (1+x)+C

a

sec

-1

x dx = x sec

-1

x- ln

|x+

x

2

1

|+C

csc

-1

x dx = x csc

-1

x+ ln

|x+

x

2

1

|+C

coth

-1

(

tanh

-1

(

x1ax

)=ln () |x| <1

a2aax

x1xa

)=ln () |x| >1

a2axa

tan

-1

(

a

x

)=

2

a

ax

2

a

x

)=

2

2

a

ax

x

)=

a

x

cot

-1

(

sec

-1

(

a

xa

22

x1

1x

2

sech()=ln(+)0≦x

2

ax

x

≦1

-1

csc

-1

(

x

)=

a

x

a

xa

22

x1

1x

2

csch ()=ln(+) |x| >0

2

ax

x

-1

页脚内容1

_x0001_

D

x

sinh x = cosh x

cosh x = sinh x

tanh x = sech

2

x

coth x = -csch

2

x

sinh x dx = cosh x + C

cosh x dx = sinh x + C

tanh x dx = ln | cosh x |+ C

coth x dx = ln | sinh x | + C

d

uv

=

u

d

v

+

v

d

u

d

uv

=

uv

=

u

d

v

=

uv

-

u

d

v

+

v

d

u

v

d

u

cos

2

θ-sin

2

θ=cos2θ

sech x = -sech x tanh sech x dx = -2tan

-1

(

e

-x

) + C cos

2

θ+ sin

2

θ=1

x

cosh

2

θ-sinh

2

θ=1

1e

x

csch x dx = 2 ln || + C

2x

csch x = -csch x coth

1e

cosh

2

θ+sinh

2

θ=cosh2θ

x

1

x

D

x

sinh()=

22

a

ax

-1

3

sin 3θ=3sinθ-4sinθ

-1-1

2

sinh x dx = x sinh x-

1x

+ C

cos3θ=4cos

3

θ-3cosθ

cosh

-1

(

1

x

)=

22

a

xa

3

cosh

-1

x dx = x cosh

-1

x-

x

2

1

+ C

→sinθ= ¼ (3sinθ-sin3θ)

tanh

-1

(

a

x

)=

2

2

a

ax

a

x

)=

2

a

ax

2

coth

-1

(

sech

-1

(

a

xax

2

x

)=

a

2

→cos

3

θ=¼(3cosθ+cos3θ)

tanh x dx = x tanh x+ ½ ln |

1-x

2

|+ C

e

jx

e

jx

sin x = cos x =

2j

coth

-1

x dx = x coth

-1

x- ½ ln |

jxjx

2

ee

1-x|+ C

2

-1-1

β

sech

-1

x dx = x sech x- sin x +

xx

ee

C

sinh x = cosh x =

2

csch

-1

x dx = x csch

-1

x+ sinh

-1

x +

e

x

e

x

C

2

-1-1

csch

-1

(

x

a

)=

a

xa

2

x

2

a

R

c

γ

b

α

正弦定理:

b

c

a

= ==2R

sin

sin

sin

余弦定理: a

2

=b

2

+c

2

-2bc cosα

b

2

=a

2

+c

2

-2ac cosβ

c

2

=a

2

+b

2

-2ab cosγ

页脚内容2


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