Questions tagged [inverse-hyperbolic-functions]

The inverse hyperbolic functions are analogous to the inverse circular trigonometric functions.

For example, as the inverse circular sine of a real number in $[-1,1]$ gives the corresponding length of the arc/angle of the sector in a unit circle whose sine is the given value, the inverse hyperbolic sine of a real number gives the area of the hyperbolic sector/hyperbolic angle of the unit hyperbola($x^2-y^2=1$) whose hyperbolic sine is the given value.

Hence, just like the inverse circular sine is denoted by $\arcsin$ besides $\sin^{-1}$, the inverse hyperbolic sine is denoted by $\text{arsinh}$ apart from $\sinh^{-1}$.

They can be defined in terms of logarithms or their corresponding circular counterparts, similar to hyperbolic functions: $$\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)=-i\sin^{-1}ix$$ $$\cosh^{-1}x=\ln\left(x+\sqrt{x^2-1}\right)=i(\text{sgn}(1-x^2))\cos^{-1}x$$ $$\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right)=-i\tan^{-1}ix$$

$$\text{csch}^{-1}x=\sinh^{-1}\frac1{x}, \text{sech}^{-1}x=\cosh^{-1}\frac1{x}, \coth^{-1}x=\tanh^{-1}\frac1{x}$$

The derivatives of $\sinh^{-1}x, \cosh^{-1}x$ and $\tanh^{-1}x$ are particularly useful to denote some standard algebraic integrals concisely and relate them with the corresponding ones whose answers involve inverse circular functions:

$$\frac{\mathrm d}{\mathrm dx}\sinh^{-1}x=\frac1{\sqrt{1+x^2}}\text{ is analogous to }\frac{\mathrm d}{\mathrm dx}\sin^{-1}x=\frac1{\sqrt{1-x^2}}$$ $$\frac{\mathrm d}{\mathrm dx}\cosh^{-1}x=\frac1{\sqrt{x^2-1}}\text{ is analogous to }\frac{\mathrm d}{\mathrm dx}\cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}$$ $$\frac{\mathrm d}{\mathrm dx}\tanh^{-1}x=\frac1{1-x^2}\text{ is analogous to }\frac{\mathrm d}{\mathrm dx}\tan^{-1}x=\frac1{1+x^2}$$

Reference : Inverse Hyperbolic Functions