[数学]三角函数与双曲函数及其导数和不定积分

三角函数与双曲函数及其导数和不定积分

定义

$$\begin{array}{rl}\sin x& =\frac{e^{ix}-e^{-ix}}{2i}\\ \cos x& =\frac{e^{ix}+e^{-ix}}{2}\\ \tan x& =\frac{e^{ix}-e^{-ix}}{(e^{ix}+e^{-ix})i}\\ \arcsin x& =-i\ln (\sqrt{1-x^2}+ix)\\ \arccos x& =-i\ln (\sqrt{x^2-1}+x)\\ \arctan x& =-\frac{i}{2}\ln (\frac{1+ix}{1-ix})\\ \mathrm{sh}x& =\frac{e^x-e^{-x}}{2}\\ \mathrm{ch}x& =\frac{e^x+e^{-x}}{2}\\ \mathrm{th}x& =\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ \text{arsh }x& =\ln (\sqrt{x^2+1}+x)\\ \text{arch }x& =\ln (\sqrt{x^2-1}+x)\\ \text{arth }x& =\frac{1}{2}\ln (\frac{1+x}{1-x})\end{array}$$

关系

$$\begin{array}{rl}\sin ix& =i\mathrm{sh}x\\ \mathrm{sh}ix& =i\sin x\\ \cos ix& =\mathrm{ch}x\\ \mathrm{ch}ix& =\cos x\\ \tan ix& =i\mathrm{th}x\\ \mathrm{th}ix& =i\tan x\\ \arcsin ix& =i\text{ arsh }x\\ \text{arsh }ix& =i\arcsin x\\ \arccos ix& =-i\text{ arch }ix\\ \text{arch }ix& =i\arccos ix\\ \arctan ix& =i\text{ arth }x\\ \text{arth }ix& =i\arctan x\end{array}$$

导数

$$\begin{array}{rl}{\left(\sin x\right)}^{\prime }& =\cos x\\ {\left(\cos x\right)}^{\prime }& =-\sin x\\ {\left(\tan x\right)}^{\prime }& =\frac{1}{{\cos }^{2}x}\\ {\left(\arcsin x\right)}^{\prime }& =\frac{1}{\sqrt{1-x^2}}\\ {\left(\arccos x\right)}^{\prime }& =-\frac{1}{\sqrt{1-x^2}}\\ {\left(\arctan x\right)}^{\prime }& =\frac{1}{1+x^2}\\ {\left(\mathrm{sh}x\right)}^{\prime }& =\mathrm{ch}x\\ {\left(\mathrm{ch}x\right)}^{\prime }& =\mathrm{sh}x\\ {\left(\mathrm{th}x\right)}^{\prime }& =\frac{1}{{\mathrm{ch}}^{2}x}\\ {\left(\text{arsh }x\right)}^{\prime }& =\frac{1}{\sqrt{x^2+1}}\\ {\left(\text{arch }x\right)}^{\prime }& =\frac{1}{\sqrt{x^2-1}}\\ {\left(\text{arth }x\right)}^{\prime }& =\frac{1}{1-x^2}\end{array}$$

不定积分

$$\begin{array}{rl}\int \sin x\text{d}x& =-\cos x+C\\ \int \cos x\text{d}x& =\sin x+C\\ \int \tan x\text{d}x& =-\ln \left|\cos x\right|+C\\ \int \mathrm{sh}x\text{d}x& =\mathrm{ch}x+C\\ \int \mathrm{ch}x\text{d}x& =\mathrm{sh}x+C\\ \int \mathrm{th}x\text{d}x& =\ln \left|\mathrm{ch}x\right|+C\end{array}$$

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注: $\text{arch }x$ 完整解析式为 $\pm \ln (\sqrt{x^2-1}+1)$, 且 $\arccos x=\pm i\text{ arch }x$.