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Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

The real inverse of the hyperbolic sine is defined on all real numbers and has derivative $1/\sqrt{1+x^2}$. It appears in arc length and in integrals involving square roots of sums of squares, where a hyperbolic substitution can simplify the expression.

For $x>0$, $\sqrt{1+x^2}$ increases from $1$, while $2/\sqrt{1+x^2}$ decreases from $2$. They meet at $(1,\sqrt{2})$, after which their order reverses. In particular, their positive separation on $0<x<1$ is $(1-x^2)/\sqrt{1+x^2}$.

Use the hyperbolic substitution $x=\sinh t$, $dx=\cosh tdt$. With $T=\mathrm{arsinh}1$, the definite integral for the area becomesSince $\sinh T=1$, $\cosh T=\sqrt{2}$ and $T=\log (1+\sqrt{2})$, the answer is

For the open matter-dominated Friedmann solution, write $\Omega ={\Omega }_{m,0}$, so $0<\Omega <1$. The present Friedmann equation gives $-K=H_0^2(1-\Omega )$. With $A=H_0^2\Omega$ and $B=H_0^2(1-\Omega )>0$, the expanding equation becomesChoose the hyperbolic substitutionThen $da=(A/(2B))\sinh \eta d\eta$ and $\sqrt{a/(a+A/B)}=\tanh (\eta /2)$. Their product givesTake the Big Bang to be $t=0$, $\eta =0$, and integrate. The parametric solution isThe parameter is determined explicitly byThus $\eta$ is conformal time multiplied by $\sqrt{-K}$, with its origin at the Big Bang; it is not equal to unscaled cosmic time. In particular today ${\eta }_0=\mathrm{arcosh}(2/\Omega -1)$. Small $\eta$ gives $a\propto {\eta }^2$ and $t\propto {\eta }^3$, hence $a\propto t^{2/3}$. Large $\eta$ gives $a/t\to H_0\sqrt{1-\Omega }=\sqrt{-K}$, in agreement with the curvature-dominated universe limit.

For the graph of a differentiable function, the arc length is $L=\int \sqrt{1+(y^{\prime }{)}^2}dx$. Since $y^{\prime }=2x$,Use the hyperbolic substitution $2x=\sinh t$: then $dx=\frac{1}{2}\cosh tdt$ and $\sqrt{1+4x^2}=\cosh t$. The hyperbolic functions ${\cosh }^{2}t=(1+\cosh 2t)/2$ yieldsHere $\mathrm{arsinh}t=\ln (t+\sqrt{1+t^2})$ is the inverse hyperbolic sine.