双曲線関数の性質

$$\begin{array}{r}\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}\\ \\ \cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}\\ \\ \tanh x={\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}}}\end{array}$$

$${\cosh }^{2}x-{\sinh }^{2}x=1$$

$$\tanh x={\displaystyle \frac{\sinh x}{\cosh x}}$$

$$1-{\tanh }^{2}x={\displaystyle \frac{1}{{\cosh }^{2}x}}$$

$$\sinh (\alpha \pm \beta )=\sinh \alpha \cosh \beta \pm \cosh \alpha \sinh \beta$$

$$\cosh (\alpha \pm \beta )=\cosh \alpha \cosh \beta \pm \sinh \alpha \sinh \beta$$

$$\tanh (\alpha \pm \beta )={\displaystyle \frac{\tanh \alpha \pm \tanh \beta }{1\pm \tanh \alpha \tanh \beta }}$$

$$\sinh 2x=2\sinh x\cosh x$$

$$\begin{array}{rl}\cosh 2x& ={\cosh }^{2}x+{\sinh }^{2}x\\ \\ & =2{\cosh }^{2}x-1\\ \\ & =1+2{\sinh }^{2}x\end{array}$$

$$\tanh 2x={\displaystyle \frac{2\tanh x}{1+{\tanh }^{2}x}}$$

$${\sinh }^{2}x={\displaystyle \frac{\cosh 2x-1}{2}}$$

$${\cosh }^{2}x={\displaystyle \frac{\cosh 2x+1}{2}}$$

$${\tanh }^{2}x={\displaystyle \frac{\cosh 2x-1}{\cosh 2x+1}}$$

$$(\sinh x{)}^{\prime }=\cosh x$$

$$(\cosh x{)}^{\prime }=\sinh x$$

$$(\tanh x{)}^{\prime }={\displaystyle \frac{1}{{\cosh }^{2}x}}$$

$${\displaystyle \int \sinh xdx=\cosh x+C}$$

$${\displaystyle \int \cosh xdx=\sinh x+C}$$

$${\displaystyle \int \tanh xdx=\log (\cosh x)+C}$$

$$\mathrm{arsinh}x=\log (x+\sqrt{x^2+1})$$

$$y=\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$$

$$x={\displaystyle \frac{e^y-e^{-y}}{2}}$$

$$e^{2y}-2x{e}^y-1=0$$

$$e^y=x+\sqrt{x^2+1}$$

$$y=\log (x+\sqrt{x^2+1})$$

$$\mathrm{arcosh}x=\log (x+\sqrt{x^2-1})$$

$$y=\cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$$

$$x={\displaystyle \frac{e^y+e^{-y}}{2}}$$

$$e^{2y}-2x{e}^y+1=0$$

$$e^y=x+\sqrt{x^2-1}$$

$$y=\log (x+\sqrt{x^2-1})$$

$$\mathrm{artanh}x={\displaystyle \frac{1}{2}\log {\displaystyle \frac{1+x}{1-x}}}$$

$$y=\tanh x={\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}}}$$

$$\begin{array}{rl}x& ={\displaystyle \frac{e^y-e^{-y}}{e^y+e^{-y}}}\\ \\ & ={\displaystyle \frac{e^{2y}-1}{e^{2y}+1}}\end{array}$$

$$e^{2y}={\displaystyle \frac{1+x}{1-x}}$$

$$2y=\log {\displaystyle \frac{1+x}{1-x}}$$

$$y={\displaystyle \frac{1}{2}\log {\displaystyle \frac{1+x}{1-x}}}$$

$${\displaystyle \int {\displaystyle \frac{1}{\sqrt{1+x^2}}dx}}$$

$$\begin{array}{rl}{\displaystyle \int {\displaystyle \frac{1}{\sqrt{1+x^2}}dx}}& ={\displaystyle \int {\displaystyle \frac{1}{\sqrt{1+{\sinh }^{2}t}}\cosh tdt}}\\ \\ & ={\displaystyle \int {\displaystyle \frac{\cosh t}{\cosh t}dt}}\\ \\ & =t+C\\ \\ & =\mathrm{arsinh}t+C\\ \\ & =\log (x+\sqrt{x^2+1})+C\end{array}$$

$${\displaystyle \int \sqrt{1+x^2}dx}$$

$$\begin{array}{rl}{\displaystyle \int \sqrt{1+x^2}dx}& ={\displaystyle \int \sqrt{1+{\sinh }^{2}t}\cosh tdt}\\ \\ & ={\displaystyle \int {\cosh }^{2}tdt}\\ \\ & ={\displaystyle \int {\displaystyle \frac{\cosh 2t+1}{2}dt}}\\ \\ & ={\displaystyle \frac{1}{4}\sinh 2t+{\displaystyle \frac{1}{2}t+C}}\\ \\ & ={\displaystyle \frac{1}{2}(\sinh t\cosh t+t)+C}\end{array}$$

$$\begin{array}{rl}\sinh t& =x\\ \\ \cosh t& =\sqrt{1-x^2}\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{1}{2}(\sinh t\cosh t+t)+C}& ={\displaystyle \frac{1}{2}(x\sqrt{1-x^2}+\mathrm{arsinh}x)+C}\\ \\ & ={\displaystyle \frac{1}{2}\{x\sqrt{1-x^2}+\log (x+\sqrt{x^2+1})\}+C}\end{array}$$

$${\displaystyle \int \sqrt{1+x^2}dx={\displaystyle \frac{1}{2}\{x\sqrt{1-x^2}+\log (x+\sqrt{x^2+1})\}+C}}$$

$$\begin{array}{r}\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}\\ \\ \cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}\\ \\ \tanh x={\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}}}\end{array}$$