微积分公式大全

2023-04-27 18:45:12 第一文档网 [ 字体：小中大 ] [

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【#第一文档网# 导语】以下是®第一文档网的小编为您整理的《微积分公式大全》，欢迎阅读！
微积分,公式,大全

微积分公式

Dx sin x=cos x cos x = -sin x tan x = sec2 x cot x = -csc2 x sec x = sec x tan x csc x = -csc x cot 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

 sec x dx = ln |sec x + tan x | + C  csc x dx = ln |csc x – cot 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-1 x csc-1(-x) = - csc-1 x

xx1

Dx sin-1 ()= sinh-1 ()= ln (x+a2x2) xR  sin-1 x dx = x sin-1 x+1x2+C

aaa2x2

 cos-1 x dx = x cos-1 x-1x2+C -1x1-1xcosh ()=ln (x+x2a2) x≧1 cos ()=

aaa2x2 tan-1 x dx = x tan-1 x-½ln (1+x2)+C

1ax-1xatanh ()=ln () |x| <1 -1x-1-12tan ()=2 a2aax cot x dx = x cot x+½ln (1+x)+C aax2

1xa-1x-1-12acoth ()=ln () |x| >1 -1x sec x dx = x sec x- ln |x+x1|+C cot ()=2 a2axa2

aax

 csc-1 x dx = x csc-1 x+ ln |x+x21|+C 11x2-1xa-1xsech()=ln(+)0≦x≦1 sec ()= 2

22axxaxxa

2

x11x-1xacsch ()=ln(+) |x| >0 csc-1 ()= 2axxaxx2a2

Dx sinh x = cosh x

cosh x = sinh x tanh x = sech2 x coth x = -csch2 x

sech x = -sech x tanh x csch x = -csch x coth 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  sech x dx = -2tan-1 (e-x) + C

1ex

 csch x dx = 2 ln || + C

2x

1e

-1

-1

2

duv = udv + vdu

 duv = uv = udv + vdu →udv = uv - vdu cos2θ-sin2θ=cos2θ cos2θ+ sin2θ=1 cosh2θ-sinh2θ=1

cosh2θ+sinh2θ=cosh2θ

3

sin 3θ=3sinθ-4sinθ  sinh x dx = x sinh x-1x+ C

cos3θ=4cos3θ-3cosθ -1-12

 cosh x dx = x cosh x-x1+ C

→sin3θ= ¼ (3sinθ-sin3θ) -1-12

 tanh x dx = x tanh x+ ½ ln | 1-x|+ C

→cos3θ=¼(3cosθ+cos3θ) -1-12

 coth x dx = x coth x- ½ ln | 1-x|+ C

ejxejxejxejx-1-1-1

sin x = cos x =  sech x dx = x sech x- sin x + C

22j

 csch-1 x dx = x csch-1 x+ sinh-1 x + C

exexexex sinh x = cosh x = γ

22

R a b

abcα

正弦定理:===2R

sinsinsinc

余弦定理: a2=b2+c2-2bc cosα

b2=a2+c2-2ac cosβ

β

c2=a2+b2-2ab cosγ

sin (α±β)=sin α cos β± cos α sin β sin α + sin β = 2 sin ½(α+β) cos ½(α-β) cos (α±β)=cos α cos β sin α sin β sin α - sin β = 2 cos ½(α+β) sin ½(α-β) 2 sin α cos β = sin (α+β) + sin (α-β) cos α + cos β = 2 cos ½(α+β) cos ½(α-β) 2 cos α sin β = sin (α+β) - sin (α-β) cos α - cos β = -2 sin ½(α+β) sin ½(α-β) 2 cos α cos β = cos (α-β) + cos (α+β)

x1

Dx sinh()=

22aax

x1

cosh-1()=

22axaax

tanh-1()= 2 2

aax

ax

coth-1()=2

aax2xasech-1()=

22axax

xacsch-1()=

22axax

-1

2 sin α sin β = cos (α-β) - cos (α+β)

x2x3xn

e=1+x+++…++ …

2!3!n!x3x5x7(1)nx2n1

sin x = x-+-+…++ …

3!5!7!(2n1)!

x

tan (α±β)=

tantancotcot

, cot (α±β)=

tantancotcot

1= n

n

i= ½n (n+1)

12

= n (n+1)(2n+1) i6i1

i1nn

i1n

x2x4x6(1)nx2n

cos x = 1-+-+…++ …

2!4!6!(2n)!x2x3x4(1)nxn1

ln (1+x) = x-+-+…++ …

234(n1)!

-1

i

i1

3

= [½n (n+1)]2

1x-1x-1-t2x-1t2x3x5x7(1)nx2n1(lnΓ(x) = e dt = 2dt = ttetan x = x-+-+…++ … 0t) dt 00357(2n1)

1

m-1n-1r(r1)r(r1)(r2)β(m, n) =x(1-x) dx=22sin2m-1x cos2n-1x dx r23

(1+x)=1+rx+x+x+… -100

2!3!m1

x= dx 0(1x)mn

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