Hyperbolic Functions and Inverse Integrals

Define the functions through exponentials

Hyperbolic sine and cosine are

\sinh x=\frac{e^x-e^{-x}}2,

\cosh x=\frac{e^x+e^{-x}}2.

Differentiating directly gives $(\sinh x)'=\cosh x$ and $(\cosh x)'=\sinh x$ . Expanding the definitions shows $\cosh^2x-\sinh^2x=1$ . Unlike ordinary cosine, hyperbolic cosine is never negative and is at least one.

The quotient $\tanh x=\sinh x/\cosh x$ has derivative $\operatorname{sech}^2x=1/\cosh^2x$ . These functions model shapes and transitions such as a hanging cable and smooth saturation, but their definitions are algebraic and do not depend on a particular application.

Visual guide

VISUAL GUIDEHyperbolic cosine and sine

cosh x
sinh x

Cosh x is even and at least 1; sinh x is odd. Their derivatives exchange the two functions. Unlike circular sine and cosine, their magnitudes grow exponentially as |x| increases.

Worked example: reverse a chain rule

For $\int\cosh(3x)dx$ , the derivative of $\sinh(3x)$ is $3\cosh(3x)$ , so the antiderivative is $\sinh(3x)/3+C$ . Likewise, $\int\operatorname{sech}^2(2x)dx=\tanh(2x)/2+C$ .

The notation $\sinh^{-1}x$ often means the inverse function, not the reciprocal. Writing $\operatorname{arsinh}x$ avoids this ambiguity. The reciprocal of $\sinh x$ is instead $\operatorname{csch}x$ .

Inverse hyperbolic sine

Since $\sinh$ is strictly increasing onto the real numbers, it has a real inverse everywhere. Implicit differentiation of $x=\sinh y$ gives

\frac d{dx}\operatorname{arsinh}x=\frac1{\sqrt{1+x^2}}.

The positive square root follows because $\cosh y>0$ . Solving the exponential definition also gives $\operatorname{arsinh}x=\ln(x+\sqrt{1+x^2})$ .

Worked example: a positive quadratic radical

For $\int dx/\sqrt{x^2+4}$ , set $u=x/2$ . Then $dx=2du$ and the denominator is $2\sqrt{1+u^2}$ . The integral is $\operatorname{arsinh}(x/2)+C$ . An equivalent logarithmic form is $\ln(x+\sqrt{x^2+4})+C$ , because the two versions differ by a constant.

Other inverse functions have narrower domains: $\operatorname{arcosh}x$ conventionally has domain $x\ge1$ and derivative $1/\sqrt{x^2-1}$ for $x>1$ ; $\operatorname{artanh}x$ has real domain $|x|<1$ and derivative $1/(1-x^2)$ . Matching a derivative formula never removes these domain choices.

Practice

Differentiate $\cosh(2x)$ .
Integrate $1/\sqrt{9+x^2}$ .
Show why $\cosh^2x+\sinh^2x=1$ is not the hyperbolic identity.

Show worked solutions

$2\sinh(2x)$ .
Scale $u=x/3$ to obtain $\operatorname{arsinh}(x/3)+C$ .
The exponential definitions give the difference of the squares as one. At nonzero $x$ , both squares are positive and their sum exceeds one.