高等数学公式汇总

等价无穷小

$\begin{aligned} & \large \sin{x} \sim \tan{x} \sim \arcsin{x} \sim \arctan{x} \sim \ln(1+x) \sim \text{e}^x-1 \sim x \\ & \large x-\sin{x} \sim \frac{1}{6}x^3 \quad\quad\quad\quad x-\arcsin{x} \sim -\frac{1}{6}x^3 \\ & \large x-\tan{x} \sim -\frac{1}{3}x^3 \quad\quad\quad\quad x-\arctan{x} \sim -\frac{1}{3}x^3 \\ & \large 1-\cos{x} \sim \frac{1}{2}x^2 \quad\quad\quad\quad 1-\cos^ax \sim \frac{a}{2}x^2 \\ & \large a^x-1 \sim x\ln a \quad\quad\quad\quad (1+x)^a-1 \sim ax \\ \end{aligned}$

泰勒公式

$\begin{aligned} & \large \sin{x}=x-\frac{x^3}{3!}+o(x^3)\quad\quad\quad\quad \arcsin{x}=x+\frac{x^3}{3!}+o(x^3)\\ & \large \tan{x}=x+\frac{x^3}{3}+o(x^3) \quad\quad\quad\quad \arctan{x}=x-\frac{x^3}{3}+o(x^3) \\ & \large \cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4) \\ & \large (1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+o(x^2)\\ & \large \text{e}^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3)\\ & \large \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)\\ \end{aligned}$

基本求导公式

$\newcommand{\arccot}{\mathrm{arccot}\,} \begin{aligned} & \large (x^a)'=ax^{a-1} \quad,\quad (\sqrt{x})' = \frac{1}{2\sqrt{x}} \quad,\quad (\frac{1}{x})' = -\frac{1}{x^2}\\ ~\\ & \large (a^x)'=a^x\ln{a} \quad,\quad (e^x)'=e^x \\ ~\\ & \large (\log_ax)'=\frac{1}{xlna} \quad,\quad (\ln{x})'=\frac{1}{x} \\ ~\\ & \large (\sin{x})' = \cos{x} \quad,\quad (\cos{x})'=-\sin{x} \quad,\quad (\tan{x})'=\sec^2{x}\\~\\ & \large (\cot{x})' = -\csc^2{x} \quad,\quad (\sec{x})'=\sec{x}\tan{x} \quad,\quad (\csc{x})'=-\csc{x}\cot{x}\\ ~\\ & \large (\arcsin{x})'=\frac{1}{\sqrt{1-x^2}} \quad,\quad (\arccos{x})'=-\frac{1}{\sqrt{1-x^2}}\\ ~\\ & \large (\arctan{x})'=\frac{1}{1+x^2} \quad,\quad (\arccot{x})'=-\frac{1}{1+x^2} \end{aligned}$

麦克劳林级数

$\begin{aligned} & \large \text{e}^x =1+x+\frac{x^2}{2}+\cdots+\frac{x^n}{n!}+o(x^n)\\ ~\\ & \large \sin{x}=x-\frac{x^3}{3!}+\cdots+\frac{(-1)^n}{(2n+1)!}x^{2n+1}+o(x^{2n+1})\\ ~\\ & \large \cos{x}=1-\frac{x^2}{2!}+\cdots+\frac{(-1)^n}{(2n)!}x^{2n}+o(x^{2n})\\ ~\\ & \large \frac{1}{1-x}=1+x+x^2+\cdots+x^n+o(x^n)\\ ~\\ & \large \frac{1}{1+x}=1-x+x^2-\cdots+(-1)^nx^n+o(x^n)\\ ~\\ & \large \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+\frac{(-1)^{n-1}}{n}x^n+o(x^n) \\ ~\\ & \large (1+x)^a=1+ax+\frac{a(a-1)}{2!}x^2+\cdots+\frac{a(a-1)\cdots(a-n+1)}{n!}x^n+o(x^n)\\ ~\\ & \large \arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots+\frac{(-1)^n}{2n+1}x^{2n+1}+o(x^{2n+1}) \end{aligned}$

不定积分基本公式

$\begin{aligned} & \large \int x^a dx = \frac{1}{a+1}x^{a+1}+C \quad,\quad \int \frac{1}{x}dx = \ln|x|+C \\ ~\\ & \large \int a^x dx = \frac{a^x}{\ln{a}}+C \quad,\quad \int \text{e}^x dx = \text{e}^x+C\\~\\ & \large \int \sin{x}dx=-\cos{x}+C\quad,\quad \int \cos{x}dx=\sin{x}+C \\ ~\\ & \large \int \tan{x}dx=-\ln|\cos{x}|+C \quad,\quad \int \cot{x}dx=\ln|\sin{x}|+C\\ ~\\ & \large \int \sec{x}dx = \ln|\sec{x}+\tan{x}|+C \quad,\quad \int \csc{x}dx=\ln|\csc{x}-\cot{x}|+C \\ ~\\ & \large \int \sec^2{x}dx=\tan{x}+C \quad,\quad \int \csc^2{x}dx=-\cot{x}+C \\ ~\\ & \large \int \sec{x}\tan{x}=\sec{x}+C \quad,\quad \int \csc{x}\cot{x}dx=-\csc{x}+C \\ ~\\ & \large \int \frac{1}{\sqrt{1-x^2}}dx=\arcsin{x}+C\quad,\quad \int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin{\frac{x}{a}}+C \\ ~\\ & \large \int \frac{1}{1+x^2}dx=\arctan{x}+C \quad,\quad \int \frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan{\frac{x}{a}}+C \\ ~\\ & \large \int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C \quad,\quad \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln|\frac{a+x}{a-x}|+C \\ ~\\ & \large \int \frac{1}{\sqrt{x^2+a^2}}dx=\ln(x+\sqrt{x^2+a^2})+C \\ ~\\ & \large \int \frac{1}{\sqrt{x^2-a^2}}dx=\ln|x+\sqrt{x^2-a^2}|+C \\ ~\\ & \large \int \sqrt{a^2-x^2}dx = \frac{a^2}{2}\arcsin{\frac{x}{a}}+\frac{x}{2}\sqrt{a^2-x^2}+C \end{aligned}$

定积分特殊性质

  设f(x)在[0,1]上连续，则

$\begin{aligned} & \large (1)\quad \int_0^\frac{\pi}{2}f(sinx)dx=\int_0^\frac{\pi}{2}f(cosx)dx \\ ~\\ & \large (2)\quad \large I_n=\int_0^\frac{\pi}{2}\sin^{n}{x}dx=\int_0^\frac{\pi}{2}\cos^{n}{x}dx \\ ~\\ & \large I_n=\frac{n-1}{n}I_{n-2} \quad,\quad I_1=1 \quad,\quad I_0=\frac{\pi}{2} \\ ~\\ & \large (3)\quad \int_0^\pi f(sinx)dx=2\int_0^\frac{\pi}{2}f(sinx)dx \\ ~\\ & \large (4)\quad \int_0^\pi xf(sinx)dx=\frac{\pi}{2}\int_0^\pi f(sinx)dx=\pi\int_0^\frac{\pi}{2}f(sinx)dx \\ ~\\ & \large (5)\quad \int_0^{2\pi} f(\vert{sinx}\vert)dx=4\int_0^\frac{\pi}{2} f(sinx)dx \\ ~\\ & \large \int_0^{2\pi} f(\vert{cosx}\vert)dx=4\int_0^\frac{\pi}{2} f(cosx)dx \\ ~\\ \end{aligned}$

  设f(x)是以T为周期的连续函数

$\begin{aligned} & \large (1)\quad \int_a^{a+T}f(x)dx=\int_0^{T}f(x)dx \\ ~\\ & \large (2)\quad \int_0^{nT}f(x)dx=n\int_0^{T}f(x)dx \\ ~\\ \end{aligned}$

  其他性质

$\begin{aligned} & \large (1)\quad \int_0^a \sqrt{a^2-x^2}dx=\frac{\pi}{4}a^2 \\ ~\\ & \large (2) \end{aligned}$

定积分的几何应用

面积

  设D由曲线y=f(x)，y=g(x)，x=a及x=b围成，则D的面积为

$\large A=\int_a^b\vert{f(x)-g(x)}\vert dx$

  设D由$\large r=r(θ)(\alpha\leq\theta\leq\beta)$围成，则D的面积为

$\large A=\frac{1}{2}\int_\alpha^\beta r^2(\theta) d\theta$

  设D由$\large r=r_1(θ),r=r_2(θ)(r_1(θ) \leq r_2(θ),\alpha\leq\theta\leq\beta)$围成，则D的面积为

$\large A=\frac{1}{2}\int_\alpha^\beta[r_2^2(\theta)-r_1^2(\theta)]d\theta$

  设$\large L:y=f(x)(a\leq x\leq b)$，则L绕x轴旋转所得旋转体侧面积为

$\large A=2\pi\int_a^b \vert{f(x)}\vert\cdot\sqrt{1+f'^2(x)}dx$

  若$\large L:\begin{cases} x=\varphi(x),\\ y=\psi(x), \end{cases} (\alpha\leq t\leq\beta)$，则L绕x轴旋转所得侧面积为

$\large A=2\pi\int_\alpha^\beta \vert{\psi(t)}\vert \cdot\sqrt{\varphi'^2(t)+\psi'^2(t)}dt$

体积

  设$\large L:y=f(x)(a\leq x\leq b)$，则L与x轴围成的图形绕x轴旋转一周所得旋转体的体积为

$\large V_x=\pi\int_a^bf^2(x)dx$

  设$\large L:y=f(x)(a\leq x\leq b)(ab\geq0)$，则L与x轴围成的图形绕y轴旋转一周所得体积为

$\large V_y=2\pi\int_a^b\vert{x}\vert\cdot\vert{f(x)}\vert dx$

曲线长度

  设$\large L:y=f(x)(a\leq x\leq b)$，则曲线L的长度为$\large l=\int_a^b \sqrt{1+f'^2(x)}dx$

  设$\large L:\begin{cases} x=\varphi(x),\\ y=\psi(x), \end{cases} (\alpha\leq t\leq\beta)$，则曲线L的长度为$\large A=\int_\alpha^\beta \sqrt{\varphi'^2(t)+\psi'^2(t)}dt$

  设$\large r=r(θ)(\alpha\leq\theta\leq\beta)$，则曲线L的长度为$\large l=\int_\alpha^\beta\sqrt{r^2(\theta)-r'^2(\theta)}d\theta$

微分方程常见形式

可分离变量

  --------略

齐次微分方程

  形式：$\large f(x,y)=\varphi(\frac{y}{x})$

  解法：令$\large \frac{y}{x}=u,\ \frac{dy}{dx}=u+x \frac{du}{dx}$

一阶齐次线性微分方程

  形式：$\large \frac{dy}{dx}+P(x)y=0$ 或 $\large y'+P(x)y=0$

  解法：通解公式：$\large y=Ce^{-\int P(x)dx}$

一阶非齐次线性微分方程

  形式：$\large \frac{dy}{dx}+P(x)y=Q(x)$ 或 $\large y'+P(x)y=Q(x)$

  解法：通解公式：$\large y=[\int{Q(x)e^{\int P(x)dx}dx+C}]\cdot e^{-\int P(x)dx}$

可降阶的高阶微分方程

  形式：$\large y^{(n)}=f(x)$

  解法：略

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  形式：$\large f(x,y',y'')=0$（缺$\large y$）

  解法：令$\large y'=p,\ y''=\frac{dp}{dx}$

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  形式：$\large f(y,y',y'')=0$（缺$x$）

  解法：令$\large y'=p,\ y''=p\frac{dp}{dy}$

二阶常系数齐次线性微分方程

  形式：$\large y''+py'+qy=0\quad$($p$,$q$为常数)

  解法：取特征方程$\large \lambda^2+p\lambda+q=0,\quad \varDelta=p^2-4q$

    若$\large \varDelta>0(\lambda_1\neq\lambda_2)$

    通解公式：$\large y=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$

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    若$\large \varDelta=0(\lambda_1=\lambda_2)$

    通解公式：$\large y=(C_1+C_2x)e^{\lambda_1x}$

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    若$\large \varDelta<0(\lambda_{1,2}=\alpha\pm i\beta)$

    通解公式：$\large y=e^{\alpha x}(C_1\cos{x}+C_2\sin{x})$

二阶常系数非齐次线性微分方程

  形式：$\large y''+py'+qy=f(x)\quad$($p$,$q$为常数)

  解法：二阶齐次通解+二阶非齐次特解

待续