等价无穷小
前提：\(x \rightarrow 0\)
\(\ln(1+x) \thicksim x \qquad x - \ln(1+x) \thicksim \dfrac{1}{2} x^2\)
\(e^x - 1 \thicksim x \qquad a^x - 1 \thicksim x \ln{a}\)
\(当\ a \neq 0,\ \ (1 + x)^a - 1 \thicksim ax \qquad (1 + x)^n - 1 \thicksim nx\)
拓展：\(当\ A(x) \rightarrow 0\ 且\ A(x)B(x) \rightarrow 0,\ [1 + A(x)]^{B(x)} - 1 \thicksim A(x)B(x)\)
\(\sin{x} \thicksim \arcsin{x} \thicksim \tan{x} \thicksim \arctan{x} \thicksim x\)
\([x - \sin{x}] \thicksim [\arcsin{x} - x] \thicksim \dfrac{1}{6} x^3\)
\([\tan{x} - x] \thicksim [x - \arctan{x}] \thicksim \dfrac{1}{3} x^3\)
\(1 - \cos{x} \thicksim \dfrac{1}{2} x^2\)
拓展：\(1 - \cos^a{x} \thicksim \dfrac{a}{2} x^2\)
导数+积分
基本公式
\((C)^{\prime} = 0\)
\(\begin{aligned} (x^a)^{\prime} = ax^{a-1} \qquad \int x^a\ \mathrm{d} x = \dfrac{1}{a + 1} x^{a + 1} + C \end{aligned}\)
\(\begin{aligned} (e^x)^{\prime} = e^x \qquad (a^x)^{\prime} = a^x \ln{a} \end{aligned}\)
\((\ln{|x|})^{\prime} = \dfrac{1}{x} \qquad (\log_{a}{x})^{\prime} = \dfrac{1}{x \ln{a}}\)
\((\sin{x})^{\prime} = \cos{x} \qquad (\cos{x})^{\prime} = -\sin{x}\)
\((\tan{x})^{\prime} = \sec^2{x} \qquad (\cot{x})^{\prime} = -\csc^2{x}\)
\(\begin{aligned} \int \tan{x}\ \mathrm{d} x = -\ln{|\cos{x}|} + C \qquad \int \cot{x}\ \mathrm{d} x = \ln{|\sin{x}|} + C \end{aligned}\)
\((\sec{x})^{\prime} = \sec{x} \tan{x} \qquad (\csc{x})^{\prime} = -\csc{x} \cot{x}\)
\(\begin{aligned} \int \sec{x}\ \mathrm{d} x = \ln{|\sec{x} + \tan{x}|} + C \qquad \int \csc{x}\ \mathrm{d} x = \ln{|\csc{x} - \cot{x}|} + C \end{aligned}\)
结合三角恒等变换公式
求 \(\begin{aligned} \int \sin^2{x}\ \mathrm{d} x \end{aligned}\)，用 \(\cos{2x} = 1 - 2 \sin^2{x}\)
求 \(\begin{aligned} \int \cos^2{x}\ \mathrm{d} x \end{aligned}\)，用 \(\cos{2x} = 2 \cos^2{x} - 1\)
求 \(\begin{aligned} \int \tan^2{x}\ \mathrm{d} x \end{aligned}\)，用 \(\tan^2{x} + 1 = \sec^2{x}\)
求 \(\begin{aligned} \int \cot^2{x}\ \mathrm{d} x \end{aligned}\)，用 \(\cot^2{x} + 1 = \csc^2{x}\)
\(\begin{aligned} (\arcsin{x})^{\prime} = -(\arccos{x})^{\prime} = \dfrac{1}{\sqrt{1 - x^2}} \qquad \int \dfrac{1}{\sqrt{a^2 - x^2}}\ \mathrm{d} x = \arcsin{\dfrac{x}{a}} + C \end{aligned}\)
\(\begin{aligned} (\arctan{x})^{\prime} = -(\mathrm{arccot}\,{x})^{\prime} = \dfrac{1}{1 + x^2} \qquad \int \dfrac{1}{a^2 + x^2}\ \mathrm{d} x = \dfrac{1}{a} \arctan{\dfrac{x}{a}} + C \end{aligned}\)
\(\begin{aligned} \int \dfrac{1}{\sqrt{a^2 \pm x^2}}\ \mathrm{d} x = \ln{|x + \sqrt{a^2 \pm x^2}|} + C \end{aligned}\)
\(\begin{aligned} \int \dfrac{1}{a^2 - x^2}\ \mathrm{d} x = \dfrac{1}{2a} \ln{\left| \dfrac{a + x}{a - x} \right|} + C \end{aligned}\)
常用快速公式
\((\dfrac{1}{x})^{\prime} = -\dfrac{1}{x^2}\)
\((\sqrt{x})^{\prime} = \dfrac{1}{2 \sqrt{x}}\)
\((\sin^2{x})^{\prime} = -(\cos^2{x})^{\prime} = \sin{2x}\)
\(|x|^{\prime} = \dfrac{x}{|x|}\)
泰勒+麦克劳林展开式
几何级数：\(\begin{aligned} \dfrac{1}{1-x} = \sum\limits_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots \qquad x \in (-1,1) \end{aligned}\)
换元得：\(\begin{aligned} \dfrac{1}{1+x} = \sum\limits_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - \cdots \qquad x \in (-1,1) \end{aligned}\)
二项式级数：\(\begin{aligned} (1+x)^a = 1 + ax + \dfrac{a(a-1)}{2!} x^2 + \cdots + \dfrac{\overbrace{a(a-1)\cdots(a-n+1)}^{n项}}{n!} x^n + \cdots \qquad x \in (-1,1) \end{aligned}\)
\(\begin{aligned} e^x = \sum\limits_{n=0}^{\infty} {\dfrac{x^n}{n!}} = 1 + x + \dfrac{x^2}{2!} + \cdots \qquad x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} \dfrac{e^x + e^{-x}}{2} = \sum\limits_{n=0}^{\infty} {\dfrac{x^{2n}}{(2n)!}} \qquad x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} \dfrac{e^x - e^{-x}}{2} = \sum\limits_{n=0}^{\infty} {\dfrac{x^{2n+1}}{(2n+1)!}} \qquad x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} \ln{(1+x)} = \sum\limits_{n=1}^{\infty} (-1)^{n-1} \dfrac{x^{n}}{n} = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots \qquad x \in (-1,1] \end{aligned}\)
换元得：\(\begin{aligned} -\ln{(1-x)} = \sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n} = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots \qquad x \in [-1,1) \end{aligned}\)
\(\begin{aligned} \sin{x} = \sum\limits_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \qquad x \in (-\infty,+\infty) \end{aligned}\)
求导得：\(\begin{aligned} \cos{x} = \sum\limits_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!} = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots \qquad x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} \arctan{x} = \sum\limits_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1} = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots \qquad x \in [-1,1] \end{aligned}\)
\(\tan{x} = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots\)
\(\arcsin{x} = x + (\dfrac{1}{2}) \dfrac{x^3}{3} + (\dfrac{1 \cdot 3}{2 \cdot 4}) \dfrac{x^5}{5} + \cdots\)
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