高等数学必背知识

Q: 等差数列前 n 项和

A: Sn=n2[2a1+(n−1)d]S_n=\frac{n}{2}[2a_1+(n-1)d]

Q: 等比数列前 n 项和

A:

Sn={na1,r=1a1(1−rn)1−r,r≠1S_n=\begin{cases} na_1, r=1 \\ \frac{a_1(1-r^n)}{1-r}, r \ne 1 \end{cases}

Q: 求

∑k=1nk\sum^n_{k=1}k

A:

n(n+1)2\frac{n(n+1)}{2}

Q:求

∑k=1nk2\sum^n_{k=1}k^2

A:

n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}

Q:求

∑k=1n1k(k+1)\sum^n_{k=1}\frac{1}{k(k+1)}

A:

nn+1\frac{n}{n+1}

Q:

sin2α+cos2αsin^2\alpha+cos^2\alpha

A: 1

Q:

1+tan2α1+tan^2\alpha

A:

sec2α=1cos2αsec^2\alpha=\frac{1}{cos^2\alpha}

Q:

1+cot2α1 + cot^2\alpha

A:

csc2α=1sin2αcsc^2\alpha=\frac{1}{sin^2\alpha}
sin(π2−α)=cosαsin(\frac{\pi}{2}-\alpha)=cos\alpha cos(π2−α)=sinαcos(\frac{\pi}{2}-\alpha)=sin\alpha tan(π2−α)=cotαtan(\frac{\pi}{2}-\alpha)=cot\alpha cot⁡(π2−α)=tanα\cot(\frac{\pi}{2}-\alpha)=tan\alpha sin(π2+α)=cosαsin(\frac{\pi}{2}+\alpha)=cos\alpha cos(π2+α)=−sinαcos(\frac{\pi}{2}+\alpha)=-sin\alpha tan(π2+α)=−cotαtan(\frac{\pi}{2}+\alpha)=-cot\alpha cot(π2+α)=−tanαcot(\frac{\pi}{2}+\alpha)=-tan\alpha sin(π−α)=sinαsin(\pi-\alpha)=sin\alpha cos(π−α)=−cosαcos(\pi-\alpha)=-cos\alpha tan(π−α)=−tanαtan(\pi-\alpha)=-tan\alpha cot(π−α)=−cotαcot(\pi-\alpha)=-cot\alpha sin(32π−α)=−cosαsin(\frac{3}{2}\pi-\alpha)=-cos\alpha sin(32π−α)=−cosαsin(\frac{3}{2}\pi-\alpha)=-cos\alpha cos(32π−α)=−sinαcos(\frac{3}{2}\pi-\alpha)=-sin\alpha tan(32π−α)=cotαtan(\frac{3}{2}\pi-\alpha)=cot\alpha cot(32π−α)=tanαcot(\frac{3}{2}\pi-\alpha)=tan\alpha sin(32π+α)=−cosαsin(\frac{3}{2}\pi+\alpha)=-cos\alpha cos(32π+α)=sinαcos(\frac{3}{2}\pi+\alpha)=sin\alpha tan(32π+α)=−cotαtan(\frac{3}{2}\pi+\alpha)=-cot\alpha cot(32π+α)=−tanαcot(\frac{3}{2}\pi+\alpha)=-tan\alpha sin(2π−α)=−sinαsin(2\pi-\alpha)=-sin\alpha cos(2π−α)=cosαcos(2\pi-\alpha)=cos\alpha tan(2π−α)=−tanαtan(2\pi-\alpha)=-tan\alpha cot(2π−α)=−cotαcot(2\pi-\alpha)=-cot\alpha

Q: 若 f(x)=f(2T-x)则

A: f(x)关于 T 对称

Q: 若 f(x)是可导偶函数则 f’(x)

A: f’(x)是奇函数

Q: 若 f(x)是可导奇函数，则 f’(x)

A: f’(x)是偶函数

Q: 若 f(x)是可导周期函数周期为 T,则 f’(x)的周期

A: T

Q: 连续的基函数原函数

A: 偶函数

Q: 连续偶函数的原函数

A: 只有一个基函数

Q: 连续偶函数周期为 T 且∫0Tf(x)dx=0\int_0^Tf(x)dx=0则其一一切原函数周期为

A: T

Q: f(x)在有限区间(a,b)内可导，且 f’(x)有界，则 f(x)

A: 在(a,b)内有界

基本求导公式

(xα)′=αxα−1(x^\alpha)'=\alpha x^{\alpha-1} (αx)′=αxln α(\alpha^x)'=\alpha^xln\ \alpha (ex)′=ex(e^x)'=e^x (logαx)′=1xln α(log_\alpha x)'=\frac{1}{xln\ \alpha} (lnx )′=1x(lnx\ )'=\frac{1}{x} (sin x)′=cosx(sin\ x)'= cos x (cos x)′=−sinx(cos\ x)' = -sinx (arcsin x)′=11−x2(arcsin\ x)' = \frac{1}{\sqrt{1-x^2}} (arccos x)′=−11−x2(arccos\ x)' = -\frac{1}{\sqrt{1-x^2}} (tan x)′=sec2 x(tan\ x)'=sec^2\ x (cot x)′=−csc2 x(cot\ x)'=-csc^2\ x (arctan x)′=11+x2(arctan\ x)'=\frac{1}{1+x^2} (arccot x)′=−11+x2(arccot\ x)'=-\frac{1}{1+x^2} (sec x)′=sec xtan x(sec\ x)'=sec\ xtan\ x (csc x)′=−csc xcot x(csc\ x)'=-csc\ xcot\ x [ln(x+x2+1)]′=1x2+1[ln(x+\sqrt{x^2+1})]'=\frac{1}{\sqrt{x^2+1}} [ln(x+x2−1)]′=1x2−1[ln(x+\sqrt{x^2-1})]'=\frac{1}{\sqrt{x^2-1}}

基本积分公式

∫xkdx=1k+1xk+1+C\int x^kdx=\frac{1}{k+1}x^{k+1}+C ∫1x2dx=−1x+C\int \frac{1}{x^2}dx=-\frac{1}{x}+C ∫1xdx=2x+C\int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+C ∫1xdx=ln∣x∣+C\int \frac{1}{x}dx = ln|x| + C ∫exdx=ex+C\int e^x dx=e^x+C ∫axdx=axln a+C\int a^x dx = \frac{a^x}{ln\ a}+C ∫sin xdx=−cos x+C\int sin\ xdx=-cos\ x + C ∫cos xdx=sin x+C\int cos\ xdx = sin\ x + C ∫tan xdx=−ln∣cos x∣+C\int tan\ xdx = -ln |cos\ x|+C ∫cot xdx=ln∣sin x∣+C\int cot\ x dx = ln |sin\ x|+C ∫sec xdx=ln∣sec x+tan x∣+C\int sec\ xdx= ln |sec\ x + tan\ x| + C ∫csc xdx=ln∣csc x−cot x∣+C\int csc\ x dx = ln |csc\ x -cot\ x| + C ∫sec2xdx=tan x+C\int sec^2 x dx=tan\ x+C ∫csc2 xdx=−cot x+C\int csc^2\ x dx =-cot\ x + C ∫sec xtan xdx=sec x+C\int sec\ xtan\ xdx = sec\ x+C ∫csc xcot xdx=−csc x+C\int csc\ xcot\ xdx=-csc\ x+C ∫11+x2dx=arctan x+C\int \frac{1}{1+x^2}dx=arctan\ x+C ∫1a2+x2dx=1aarctanxa+C\int \frac{1}{a^2+x^2}dx=\frac{1}{a}arctan\frac{x}{a}+C ∫11−x2dx=arcsin x+C\int \frac{1}{\sqrt{1-x^2}}dx=arcsin\ x+C ∫1a2−x2dx=arcsin1a+C\int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin\frac{1}{a}+C ∫1x2+a2dx=ln(x+x2+a2)+C\int \frac{1}{\sqrt{x^2+a^2}}dx=ln(x+\sqrt{x^2+a^2}) + C ∫1x2−a2dx=ln∣x+x2−a2∣+C\int \frac{1}{\sqrt{x^2-a^2}}dx=ln|x+\sqrt{x^2-a^2}|+C ∫1x2−a2dx=12aln∣x−ax+a∣+C\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}ln|\frac{x-a}{x+a}|+C ∫1a2−x2dx=12aln∣x+ax−a∣+C\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln|\frac{x+a}{x-a}|+C ∫a2−x2dx=a22arcsinxa+x2a2−x2+C\int \sqrt{a^2-x^2} dx = \frac{a^2}{2} arcsin \frac{x}{a} + \frac{x}{2}\sqrt{a^2-x^2} + C ∫sin2 xdx=x2−sin 2x4+C(sin2x=1−cos 2x2)\int sin^2\ x dx = \frac{x}{2} - \frac{sin\ 2x}{4} + C(sin^2x=\frac{1-cos\ 2x}{2}) ∫cos2 xdx=x2+sin 2x4+C(cos2x=1−+cos 2x2)\int cos^2\ x dx = \frac{x}{2} + \frac{sin\ 2x}{4} + C(cos^2x=\frac{1-+cos\ 2x}{2}) ∫tan2xdx=tan x−x+C(tan2x=sec2x−1)\int tan^2 x dx = tan\ x-x+C(tan^2x=sec^2x-1) ∫cot2xdx=−cot x−x+C(cot2x=csc2x−1)\int cot^2 x dx = -cot\ x-x+C(cot^2x=csc^2x-1)
sin 0=0sin\ 0 = 0 sin⁡π6=12\sin\frac{\pi}{6} = \frac{1}{2} sin⁡π4=22\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} sin⁡π3=32\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} sin⁡π2=1\sin\frac{\pi}{2} = 1 sin⁡ π=0\sin\ \pi = 0 sin⁡3π2=−1\sin\frac{3\pi}{2} = -1 sin⁡ 2π=0\sin\ 2\pi = 0 cos 0=1cos\ 0 = 1 cosπ6=32cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} cosπ4=22cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} cosπ3=12cos\frac{\pi}{3} = \frac{1}{2} cosπ2=0cos\frac{\pi}{2} = 0 cos π=−1cos\ \pi = -1 cos3π2=0cos \frac{3\pi}{2} = 0 cos⁡2π=1\cos2\pi = 1 tan⁡ 0=0\tan\ 0 = 0 tan⁡π6=33\tan\frac{\pi}{6} = \frac{\sqrt{3}}{3} tan⁡π4=1\tan\frac{\pi}{4} = 1 tan⁡π3=3\tan\frac{\pi}{3} = \sqrt{3} lim⁡x→∞tan⁡x=∞\lim_{x \to \infty}\tan x = \infty tan⁡π=0\tan\pi = 0 lim⁡x→3π2tan⁡x=∞\lim_{x \to \frac{3\pi}{2}} \tan x = \infty tan⁡2π=0\tan 2\pi = 0 lim⁡x→0cot⁡x=∞\lim_{x \to 0} \cot x = \infty cot⁡π6=3\cot \frac{\pi}{6} = \sqrt{3} cot⁡π4=1\cot\frac{\pi}{4} = 1 cot⁡π3=33\cot\frac{\pi}{3} = \frac{\sqrt{3}}{3} cot⁡π2=0\cot\frac{\pi}{2} = 0 lim⁡x→πcot⁡x=∞\lim_{x \to \pi} \cot x = \infty cot⁡3π2=0\cot\frac{3\pi}{2} = 0 lim⁡x→2πcot⁡x=∞\lim_{x \to 2\pi}\cot x = \infty arcsin⁡x+arccos⁡x=π2\arcsin x + \arccos x = \frac{\pi}{2} arcsin⁡0=0\arcsin 0 = 0 arcsin⁡12=π6\arcsin\frac{1}{2} = \frac{\pi}{6} arcsin⁡22=π4\arcsin\frac{\sqrt{2}}{2} = \frac{\pi}{4} arcsin⁡32=π3\arcsin\frac{\sqrt{3}}{2} = \frac{\pi}{3} arcsin⁡1=π2\arcsin 1 = \frac{\pi}{2} arccos⁡1=0\arccos 1 = 0 arccos⁡32=π6\arccos\frac{\sqrt{3}}{2} = \frac{\pi}{6} arccos⁡22=π4\arccos\frac{\sqrt{2}}{2} = \frac{\pi}{4} arccos⁡12=π3\arccos\frac{1}{2} = \frac{\pi}{3} arccos⁡0=π2\arccos 0 = \frac{\pi}{2} arctan⁡x+arccotx=π2\arctan x + \textrm{arccot} x = \frac{\pi}{2} arctan⁡0=0\arctan 0 = 0 arctan⁡33=π6\arctan\frac{\sqrt{3}}{3} = \frac{\pi}{6} arctan⁡1=π4\arctan 1 = \frac{\pi}{4} arctan⁡3=π3\arctan\sqrt{3} = \frac{\pi}{3} arccot0=π2\textrm{arccot} 0 = \frac{\pi}{2} arccot3=π6\textrm{arccot}\sqrt{3} = \frac{\pi}{6} arccot1=π4\textrm{arccot}1 = \frac{\pi}{4} arccot33=π3\textrm{arccot}\frac{\sqrt{3}}{3} = \frac{\pi}{3} lim⁡x→−∞arctan⁡x=−π2\lim_{x \to -\infty}\arctan x = -\frac{\pi}{2} lim⁡x→+∞arctan⁡x=π2\lim_{x \to +\infty}\arctan x = \frac{\pi}{2} lim⁡x→−∞arccotx=π\lim_{x \to -\infty}\textrm{arccot}x = \pi lim⁡x→+∞arccotx=0\lim_{x \to +\infty}\textrm{arccot}x = 0