以下仅是个人总结仅供参考(不包含微分方程模块)

\begin{aligned} \sin \alpha \cos \beta&=\frac{1}{2}[\sin (\alpha+\beta)+\sin(\alpha-\beta)] \\ \cos \alpha \sin \beta&=\frac{1}{2}[\sin (\alpha+\beta)-\sin(\alpha-\beta)] \\ \cos \alpha \cos \beta&=\frac{1}{2}[\cos (\alpha+\beta)+\cos(\alpha-\beta)] \\ \sin \alpha \sin \beta&=-\frac{1}{2}[\cos (\alpha+\beta)-\cos(\alpha-\beta)]\end{aligned}

\begin{aligned} \sin\alpha+\sin\beta&=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \sin\alpha-\sin\beta&=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} \\ \cos\alpha+\cos\beta&=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \cos\alpha-\cos\beta&=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\\ \tan\alpha+\tan\beta&=\frac{\sin (\alpha+\beta)}{\cos\alpha\cdot\cos \beta}\end{aligned}

\begin{aligned} \label{gyhgs} \sin^2 x+\cos^2x&=1\\ \sec^2 x-\tan^2x&=1\\\cosh^2x-\sinh^2x&=1\end{aligned}

\begin{aligned} \sin^2x&=\frac{1}{2}(1-\cos 2x) \\\cos^2x&=\frac{1}{2}(1+\cos 2x) \\ \tan^2x&=\frac{1-\cos 2x}{1+\cos 2x} \\ \sin x&=2\sin\frac{x}{2}\cos\frac{x}{2} \\ \cos x&=2\cos^2\frac{x}{2}-1=1-2\sin^2\frac{x}{2}=\cos^2\frac{x}{2}-\sin^2\frac{x}{2} \\ \tan x&=\frac{2\tan(x/2)}{1-\tan^2(x/2)}\end{aligned}

令 u=\tan\dfrac{x}{2} 则

\begin{aligned} \sin x=\frac{2u}{1+u^2}\\ \cos x=\frac{1-u^2}{1+u^2}\end{aligned}

\begin{aligned} \mathrm{e}^{x}&=1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\cdots+\frac{1}{n!}x^{n}+o(x^{n})\\ \ln(x+1)&=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\cdots+(-1)^{n-1}\frac{1}{n}x^{n}+o(x^{n})\end{aligned}

令 n=2m 有,

\begin{aligned} \sin x&=x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}+\cdots+(-1)^{m-1}\frac{1}{(2m-1)!}x^{2m-1}+o(x^{2m}) \\ \cos x&=1-\frac{1}{2}x^2+\frac{1}{24}x^4-\cdots+(-1)^m \frac{1}{(2m)!}x^{2m}+o(x^{2m+1}) \\ \tan x&=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\cdots+o(x^{2m-1})\end{aligned} \begin{aligned} \arcsin x&=x+\frac{1}{6}x^3+\frac{3}{40}x^{5}+\cdots+o(x^{2m})\end{aligned}

常用于近似计算的泰勒公式

\begin{aligned} \frac{1}{1-x}&=1+x+x^2+x^3+\cdots+x^n+o(x^n) \\ (1+x)^{\alpha}&=\sum_{i=0}^{n}\frac{\prod_{j=0}^{i-1}{(\alpha-j})}{i!}x^n+o(x^n)\notag \\ &=1+\alpha x+\frac{\alpha(\alpha-1)}{2}x^2+\cdots+o(x^n) \\ \alpha^x&=\sum_{i=0}^{n}\frac{\ln^n \alpha}{n!}x^n+o(x^n)\notag \\ &=1+x\ln \alpha+\frac{\ln^2 \alpha}{2}x^2+\cdots+\frac{\ln^n \alpha}{n!}x^n+o(x^n)\end{aligned}

基本求导公式

\begin{equation} \left( C\right)'=0 \\ \left( x^{\mu}\right)'=\mu x^{\mu-1} \\ \left( \sin x\right)'=\cos x \\ \left( \cos x\right)'=-\sin x \\ \left( \tan x\right)'=\sec^2 x\\ \left( \cot x\right)'=-\csc^2 x \\ \left( \sec x\right)'=\sec x\cdot\tan x \\ \left( \csc x\right)'=-\csc x\cdot\cot x \\ \left( a^x\right)'=a^x\ln a\ (a>0,a\neq1)\\ \left( \log_{a}x\right)'=\frac{1}{x\cdot\ln a}\ (a>0,a\neq1) \\ \left( \arcsin x\right)'=\frac{1}{\sqrt{1-x^2}} \\ \left( \arccos x\right)'=-\frac{1}{\sqrt{1-x^2}} \\ \left( \arctan x\right)'=\frac{1}{1+x^2} \\ \left( \mathrm{arccot}\, x\right)'=-\frac{1}{1+x^2} \\ \end{equation}

由定义式我们可以推得

曲线在对应点 M(x,y) 的曲率中心 D(\alpha,\beta) 的坐标为

\begin{cases} \alpha=x-\displaystyle\frac{y'(1+y^{'2})}{y^{''2}} \\ \beta=y+\displaystyle\frac{1+y^{'2}}{y''} \end{cases}

曲线的渐近线

\begin{aligned} &\int k \,\mathrm{d}x=kx+C \ \mbox{(其中}k\mbox{为常数)} \\ &\int x^\mu\,\mathrm{d}x=\frac{x^{\mu+1}}{\mu+1}+C\ (\mu\neq-1) \\ &\int \frac{1}{x}\,\mathrm{d}x=\ln|x|+C \\ &\int\frac{\mathrm{d}x}{1+x^2}=\arctan x+C \\ &\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin x+C_1=-\arccos x+C_2 \\ &\int \sin x\,\mathrm{d}x=-\cos x+C \\ &\int\cos x \,\mathrm{d}x=\sin x +C \\ &\int\tan x\,\mathrm{d}x=-\ln |\cos x|+C \\ &\int\cot x\,\mathrm{d}x=\ln |\sin x|+C \\ &\int\csc x\,\mathrm{d}x=\int\frac{1}{\sin{x}}\,\mathrm{d}x=\frac{1}{2} \ln{\left|\frac{1-\cos{x}}{1+\cos{x}}\right|}+C=\ln{\left|\tan{\frac{x}{2}}\right|}+C=\ln{\left|\csc{x}-\cot{x}\right|}+C \\ &\int\sec x\,\mathrm{d}x=\int\frac{1}{\cos{x}}\,\mathrm{d}x=\frac{1}{2} \ln{\left|\frac{1+\sin{x}}{1-\sin{x}}\right|}+C=\ln{\left|\sec{x}+\tan{x}\right|}+C \\ &\int\sec^2 x\,\mathrm{d}x=\tan x +C \\ &\int \csc^2 x\,\mathrm{d}x=-\cot x +C \\ &\int \sec x\cdot\tan x \,\mathrm{d}x=\sec x+C \\ &\int\csc x \cdot\cot x \,\mathrm{d}x=-\csc x+C \\ &\int \mathrm{e}^x \,\mathrm{d}x=\mathrm{e}^x+C \\ &\int a^x\,\mathrm{d}x=\frac{a^x}{\ln a}+C \\ &\int \sinh x\,\mathrm{d}x=\cosh x+C \\ &\int \cosh x\,\mathrm{d}x=\sinh x+C \\ &\int \frac{1}{a^2+x^2}\,\mathrm{d}x=\frac{1}{a}\arctan\frac{x}{a}+C \\ &\int \frac{1}{a^2-x^2}\,\mathrm{d}x=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C \\ &\int \frac{1}{\sqrt{a^2-x^2}}\,\mathrm{d}x=\arcsin\frac{x}{a}+C \\ &\int \frac{1}{\sqrt{x^2\pm a^2}}\,\mathrm{d}x=\ln \left|x+\sqrt{x^2\pm a^2}\right|+C \end{aligned}

第一类换元法

\begin{aligned} &\int {f( ax + b){\rm{d}}x = }\frac{1}{a}\int {f(ax+b){\mathrm{d}}(ax + b)\;(a \neq 0)} \\ &\int {f(a{x^{m + 1}} + b){x^m}{\rm{d}}x} = \frac{1}{{a(m + 1)}}\int {f(a{x^{m + 1}} + b){\rm{d}}(a{x^{m + 1}} + b)} \\ &\int {f\left( \frac{1}{x}\right) \frac{{{\rm{d}}x}}{{{x^2}}}\;} = - \int {f\left( \frac{1}{x}\right) {\rm{d}}\left( \frac{{\rm{1}}}{x}\right) \;} \\ &\int {f(\ln x)\frac{1}{x}} {\rm{d}}x = \int {f(\ln x){\rm{d(}}\ln x)} \\ &\int {f({\mathrm{e}^x})} {\mathrm{e}^x}{\rm{d}}x = \int {f({\mathrm{e}^x}} ){\rm{d(}}{\mathrm{e}^x}) \\ &\int {f(\sqrt x } )\frac{{{\rm{d}}x}}{{\sqrt x }} = 2\int {f(\sqrt x } ){\rm{d}}(\sqrt x ) \\ &\int {f(\sin x)\cos x{\rm{d}}x = } \int {f(\sin x){\rm{d}}\sin x} \\ &\int {f(\cos x)\sin x{\rm{d}}x = } - \int {f(\cos x){\rm{d}}\cos x} \\ &\int {f(\tan x){{\sec }^2}} x{\rm{d}}x = \int {f(\tan x){\rm{d}}\tan x} \\ &\int {f(\cot x){{\csc }^2}} x{\rm{d}}x = - \int {f(\cot x){\rm{d}}\cot x} \\ &\int {f(\arcsin x)\frac{1}{{\sqrt {1 - {x^2}} }}} {\rm{d}}x = \int {f(\arcsin x){\rm{d}}\arcsin x} \\ &\int {f(\arctan x)\frac{1}{{1 + {x^2}}}} {\rm{d}}x = \int {f(\arctan x){\rm{d}}\arctan x} \\ &\int {\frac{{f'(x)}}{{f(x)}}} {\rm{d}}x = \int {\frac{{{\rm{d}}f(x)}}{{f(x)}}} = \ln \left| f(x)\right| + C\end{aligned}

部分分式

\begin{aligned} \frac{{P(x)}}{{Q(x)}} =& \frac{{{A_1}}}{{{{(x - a)}^\alpha }}} + \frac{{{A_2}}}{{{{(x - a)}^{\alpha - 1}}}} + \cdots + \frac{{{A_\alpha }}}{{x - a}} + \notag\ \\&\frac{{{B_1}}}{{{{(x - b)}^\beta }}} + \frac{{{B_2}}}{{{{(x - b)}^{\beta - 1}}}} + \cdots + \frac{{{B_\beta }}}{{x - b}} + \notag\ \\&\frac{{{M_1}x + {N_1}}}{{{{({x^2} + px + q)}^\lambda }}} + \frac{{{M_2}x + {N_2}}}{{{{({x^2} + px + q)}^{\lambda - 1}}}} + \cdots + \frac{{{M_\lambda }x + {N_\lambda }}}{{{x^2} + px + q}} + \notag\ \\&\cdots \end{aligned}

三角函数的特殊定积分

\begin{aligned} I_n&=\int_0^{\frac{\pi}{2}}\sin^nx\,\mathrm{d}x=\int_0^{\frac{\pi}{2}}\cos^nx\,\mathrm{d}x\notag \ I_n&\\&=\frac{n-1}{n}I_{n-2}\notag\ \\&=\begin{cases} \ \dfrac{{n - 1}}{n} \cdot \dfrac{{n - 3}}{{n - 2}} \cdots \dfrac{4}{5} \cdot \dfrac{2}{3}\quad (n\mbox{为大于}1\mbox{的正奇数}),I_1=1\\ \ \dfrac{{n - 1}}{n} \cdot \dfrac{{n - 3}}{{n - 2}} \cdots \dfrac{3}{4} \cdot \dfrac{1}{2} \cdot \dfrac{\pi }{2}\quad (n\mbox{为正偶数}),I_0=\dfrac{\pi}{2} \end{cases}\end{aligned}