Inverse hyperbolic functionsEdexcel A-Level Further Maths: Revision notes

The hyperbolic functions $\sinh x$, $\cosh x$ and $\tanh x$ have inverses once each is restricted to a one-to-one domain:

• $\operatorname{arsinh}x$: domain all real $x$, range all real $y$ (the inverse of $\sinh$, which is one-to-one on $\mathbb{R}$).
• $\operatorname{arcosh}x$: domain $x\ge1$, range $y\ge0$. $\cosh$ is many-to-one, so it is restricted to $x\ge0$ before inverting; its values are all $\ge1$.
• $\operatorname{artanh}x$: domain $-1<x<1$, range all real $y$, because $\tanh x$ always lies strictly between $-1$ and $1$. The graph of each inverse is the reflection of the (restricted) hyperbolic graph in the line $y=x$. Note that $\operatorname{arsinh}$ and $\operatorname{artanh}$ are odd functions, while $\operatorname{arcosh}$ takes no negative values.

The inverse hyperbolic functions can be written as natural logarithms: $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right),\qquad\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right)\ (x\ge1),$ $\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x}\ (-1<x<1).$ These are in the formulae booklet, but you must be able to derive them. Examples: $\operatorname{arsinh}\frac34=\ln\left(\frac34+\frac54\right)=\ln2$; $\operatorname{arcosh}\frac{13}5=\ln\left(\frac{13}5+\frac{12}5\right)=\ln5$; $\operatorname{artanh}\frac13=\frac12\ln\frac{4/3}{2/3}=\frac12\ln2$.

Method: write $y=\operatorname{arsinh}x$, so $x=\sinh y=\frac{e^y-e^{-y}}2$. Multiply by $2e^y$: $e^{2y}-2xe^y-1=0$, a quadratic in $e^y$. $e^y=x\pm\sqrt{x^2+1}.$ Since $e^y>0$ and $x-\sqrt{x^2+1}<0$, reject the minus sign: $y=\ln\left(x+\sqrt{x^2+1}\right)$. arcosh: the same steps give $e^y=x\pm\sqrt{x^2-1}$. Both roots are positive (their product is 1), but $y\ge0$ needs $e^y\ge1$, so take the plus sign. artanh: $x=\tanh y=\frac{e^{2y}-1}{e^{2y}+1}$, so $e^{2y}(1-x)=1+x$ and $y=\frac12\ln\frac{1+x}{1-x}$.

Convert an inverse hyperbolic function into an exact logarithm, or solve equations by first finding a hyperbolic value. Example: solve $2\cosh^2x-3\sinh x=4$. Use $\cosh^2x=1+\sinh^2x$: $2\sinh^2x-3\sinh x-2=0$, so $\sinh x=2$ or $-\frac12$. Then $x=\operatorname{arsinh}2=\ln(2+\sqrt5)$ or $x=\operatorname{arsinh}\left(-\frac12\right)=\ln\frac{\sqrt5-1}{2}$. Equations can also be solved by writing $\sinh x$ and $\cosh x$ in terms of $e^x$ to get a quadratic in $e^x$, then taking logarithms. Keep answers exact unless told otherwise.

Two standard results (in the formulae booklet): $\int\frac{1}{\sqrt{x^2+a^2}}\,dx=\operatorname{arsinh}\frac xa+c,\qquad\int\frac{1}{\sqrt{x^2-a^2}}\,dx=\operatorname{arcosh}\frac xa+c\ (x>a).$ Substitution: use $x=a\sinh u$ (then $dx=a\cosh u\,du$ and $\sqrt{x^2+a^2}=a\cosh u$) for the first, and $x=a\cosh u$ (then $\sqrt{x^2-a^2}=a\sinh u$) for the second. The integrand collapses to $\int1\,du$. Example: $\int_0^4\frac1{\sqrt{x^2+9}}\,dx=\left[\operatorname{arsinh}\frac x3\right]_0^4=\ln\left(\frac43+\frac53\right)=\ln3$. Associated integrals: complete the square first. $\int\frac{1}{\sqrt{x^2+2x+5}}\,dx=\int\frac1{\sqrt{(x+1)^2+4}}\,dx=\operatorname{arsinh}\frac{x+1}{2}+c$.