21. Antiderivatives, Areas and the FTC
b. Antiderivative Rules
4. Hyperbolic Trig Functions (Optional)
The tables on the previous pages listed the antiderivatives of the standard special functions. The tables on this page, list the properties of the hyperbolic trig functions. These were defined in the PreCalculus chapter on Exponentials and Logarithms. Their derivatives were found in the chapter on Derivative Rules.
Definitions and Derivatives
We first review the definitions and derivatives. The derivatives can be derived by differentiating the exponential formulas in the definitions.
| Definition | Derivative | |
|---|---|---|
| Hyperbolic Sine | \(\sinh x=\dfrac{e^x-e^{-x}}{2}\) | \(\dfrac{d}{dx}\sinh x=\cosh x\) |
| Hyperbolic Cosine | \(\cosh x=\dfrac{e^x+e^{-x}}{2}\) | \(\dfrac{d}{dx}\cosh x=\sinh x\) |
| Hyperbolic Tangent | \(\tanh x=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\dfrac{\sinh x}{\cosh x}\) | \(\dfrac{d}{dx}\tanh x=\mathrm{sech}^2 x\) |
| Hyperbolic Cotangent | \(\coth x=\dfrac{e^x+e^{-x}}{e^x-e^{-x}}=\dfrac{\cosh x}{\sinh x}\) | \(\dfrac{d}{dx}\coth x=-\mathrm{csch}^2 x\) |
| Hyperbolic Secant | \(\mathrm{sech}\,x=\dfrac{2}{e^x+e^{-x}}=\dfrac{1}{\cosh x}\) | \(\dfrac{d}{dx}\mathrm{sech}\,x=-\,\mathrm{sech}\,x\tanh x\) |
| Hyperbolic Cosecant | \(\mathrm{csch}\,x=\dfrac{2}{e^x-e^{-x}}=\dfrac{1}{\sinh x}\) | \(\dfrac{d}{dx}\mathrm{csch}\,x=-\,\mathrm{csch}\,x\coth x\) |
Notice that the derivatives are identical to those for the ordinary circular trig functions except for a few minus sign differences.
We now look at the inverse hyperbolic functions. These were defined in the PreCalculus chapter on Exponentials and Logarithms. We first need to check which functions are \(1-1\), so the inverse can be defined on the whole domain. If not, then we need to pick a branch on which to define the inverse. The functions \(\mathrm{sinh}\), \(\mathrm{tanh}\), \(\mathrm{coth}\) and \(\mathrm{csch}\) are 1-1 because their derivatives are always positive or always negative, so that the functions are strictly increasing or strictly decreasing. (The functions \(\mathrm{coth}\) and \(\mathrm{csch}\) are also split at \(x=0\) but the two pieces don't overlap.) So the inverses of these four functions are well-defined. The functions \(\mathrm{cosh}\), \(\mathrm{sech}\) are not 1-1 because their derivatives change sign at \(x=0\). By definition, we pick the branch with \(x \ge 0\) to define the inverse.
Since there are exponential definitions of the hyperbolic functions, it is not surprising there are logarithmic formulas for their inverses. See the PreCalculus chapter on Exponentials and Logarithms.
The inverse hyperbolic functions are given by:
\[\begin{aligned}
&\mathrm{arcsinh}\,x=\ln(x+\sqrt{x^2+1})
&&\mathrm{arccosh}\,x=\ln(x+\sqrt{x^2-1}) \\
&\mathrm{arctanh}\,x=\dfrac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right)
&&\mathrm{arccoth}\,x=\dfrac{1}{2}\ln\left(\dfrac{x+1}{x-1}\right) \\
&\mathrm{arcsech}\,x=\ln\left(\dfrac{1+\sqrt{1-x^2}}{x}\right)
&&\mathrm{arccsch}\,x=\ln\left(\dfrac{1+\sqrt{1+x^2}}{x}\right)
\end{aligned}\]
Derivations for:
The derivations for \(\mathrm{arccosh}\), \(\mathrm{arccoth}\) and
\(\mathrm{arccsch}\) are left to the exercises.
Derivation of the Formula for the Inverse Hyperbolic Sine
To say \(\mathrm{arcsinh}\) is the inverse of \(\sinh\) means: \(b=\mathrm{arcsinh}\,a\) if and only if \(a=\sinh b=\dfrac{e^b-e^{-b}}{2}\) We solve the latter equation for \(b\): \[\begin{aligned} 2a=&e^b-e^{-b}=e^b-\dfrac{1}{e^b} \\ &2ae^b=e^{2b}-1 \\ (e^b)&^2-2a(e^b)-1=0 \end{aligned}\] This is a quadratic equation for \(e^b\). By the quadratic formula: \[ e^b=\dfrac{2a\pm\sqrt{4a^2+4}}{2}=a\pm\sqrt{a^2+1} \] We pick the \(+\) sign (so that \(e^b\) is positive) and take the log: \[ b=\ln(a+\sqrt{a^2+1})\equiv\mathrm{arcsinh}\,a \] So the formula for \(\mathrm{arcsinh}\,x\) is: \[ \mathrm{arcsinh}\,x=\ln(x+\sqrt{x^2+1}) \]
Derivation of the Formula for the Inverse Hyperbolic Tangent
To say \(\mathrm{arctanh}\) is the inverse of \(\tanh\) means: \(b=\mathrm{arctanh}\,a\) if and only if \(a=\tanh b=\dfrac{e^b-e^{-b}}{e^b+e^{-b}}\) We solve the latter equation for \(b\): \[\begin{aligned} a(e^b+e^{-b})&=e^b-e^{-b} \\ a(e^{2b}+1)&=e^{2b}-1 \\ (a-1)e^{2b}&=-1-a \\ e^{2b}&=\dfrac{-1-a}{a-1}=\dfrac{1+a}{1-a} \\ b&=\dfrac{1}{2}\ln\left(\dfrac{1+a}{1-a}\right)\equiv\mathrm{arctanh}\,a \end{aligned}\] So the formula for \(\mathrm{arctanh}\,x\) is: \[ \mathrm{arctanh}\,x=\dfrac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right) \]
Derivation of the Formula for the Inverse Hyperbolic Secant
To say \(\mathrm{arcsech}\) is the inverse of \(\mathrm{sech}\) means: \(b=\mathrm{arcsech}\,a\) if and only if \(a=\mathrm{sech}\,b=\dfrac{2}{e^b+e^{-b}}\) Before solving for \(b\), we first note that \(\mathrm{sech}\,b\) is not 1-1 since \(b=u\) and \(b=-u\) give the same value of \(\mathrm{sech}\,b\). We make the choice to define \(\mathrm{arcsech}\,a\) for the branch of \(\mathrm{sech}\,b\) with \(b \ge 0\). We now solve the latter equation for \(b\), expecting two solutions from which we want to pick the one with \(b \ge 0\): \[\begin{aligned} a(e^b+e^{-b})&=2 \\ a(e^{2b}+1)&=2e^b \\ a(e^b)^2-2(e^b)+a&=0 \\ \end{aligned}\] This is a quadratic equation for \(e^b\). By the quadratic formula: \[ e^b=\dfrac{2\pm\sqrt{4-4a^2}}{2a}=\dfrac{1\pm\sqrt{1-a^2}}{a} \] The \(+\) and \(-\) signs give the two solutions and we want the one with \(b \ge 0\). Now if \(b \ge 0\), then \(e^b \ge e^{-b}\). So we need the larger solution which is the one with the \(+\) sign. Then we take the log: \[ b=\ln\left(\dfrac{1+\sqrt{1-a^2}}{a}\right)\equiv\mathrm{arcsech}\,a \] So the formula for \(\mathrm{arcsech}\,x\) is: \[ \mathrm{arcsech}\,x=\ln\left(\dfrac{1+\sqrt{1-x^2}}{x}\right) \]
The formulas for the derivatives of the inverse hyperbolic functions can be derived by differentiating their logarithmic formulas. This was previously done in the chapter on Inverse Functions.
| Formula | Derivative | |
|---|---|---|
| Inverse Hyperbolic Sine | \(\mathrm{arcsinh}\,x=\ln(x+\sqrt{x^2+1})\) | \(\dfrac{d}{dx}\mathrm{arcsinh}\,x=\dfrac{1}{\sqrt{x^2+1}}\) |
| Inverse Hyperbolic Cosine | \(\mathrm{arccosh}\,x=\ln(x+\sqrt{x^2-1})\) | \(\dfrac{d}{dx}\mathrm{arccosh}\,x=\dfrac{1}{\sqrt{x^2-1}}\) |
| Inverse Hyperbolic Tangent | \(\mathrm{arctanh}\,x=\dfrac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right)\) | \(\dfrac{d}{dx}\mathrm{arctanh}\,x=\dfrac{1}{x^2-1}\) |
| Inverse Hyperbolic Cotangent | \(\mathrm{arccoth}\,x=\dfrac{1}{2}\ln\left(\dfrac{x+1}{x-1}\right)\) | \(\dfrac{d}{dx}\mathrm{arccoth}\,x=\dfrac{-1}{x^2-1}\) |
| Inverse Hyperbolic Secant | \(\mathrm{arcsech}\,x=\ln\left(\dfrac{1+\sqrt{1-x^2}}{x}\right)\) | \(\dfrac{d}{dx}\mathrm{arcsech}\,x=\dfrac{-1}{\sqrt{x^2(1-x^2)}}\) |
| Inverse Hyperbolic Cosecant | \(\mathrm{arccsch}\,x=\ln\left(\dfrac{1+\sqrt{1+x^2}}{x}\right)\) | \(\dfrac{d}{dx}\mathrm{arccsch}\,x=\dfrac{-1}{\sqrt{x^2(1+x^2)}}\) |
Notice that the derivatives of the inverse hyperbolic trig functions are the same as those for the ordinary inverse trig functions except for numerous minus sign differences.
Antiderivatives
We are now ready for the antiderivatives. You can check each line of the table by differentiating the quantity on the right to see you get the quantity on the left.
| If the Function is | then the Antiderivative is | |
|---|---|---|
| Hyperbolic Sine | \(f(x)=\cosh x\) | \(F(x)=\sinh x+C\) |
| Hyperbolic Cosine | \(f(x)=\sinh x\) | \(F(x)=\cosh x+C\) |
| Hyperbolic Tangent\(^\text{1}\) | \(f(x)=\mathrm{sech}^2 x\) | \(F(x)=\tanh x+C\) |
| Hyperbolic Cotangent\(^\text{1}\) | \(f(x)=\mathrm{csch}^2 x\) | \(F(x)=-\coth x+C\) |
| Hyperbolic Secant\(^\text{1}\) | \(f(x)=\mathrm{sech}\,x\,\mathrm{tanh}\,x\) | \(F(x)=-\,\mathrm{sech}\,x+C\) |
| Hyperbolic Cosecant\(^\text{1}\) | \(f(x)=\mathrm{csch}\,x\,\mathrm{coth}\,x\) | \(F(x)=-\,\mathrm{csch}\,x+C\) |
\(^\text{1}\) These are the functions whose antiderivatives are the hyperbolic trig functions: \(\tanh x\), \(\coth x\), \(\mathrm{sech}\,x\) and \(\mathrm{csch}\,x\). We will discuss the antiderivatives of these hyperbolic trig functions in Calculus 2 in the chapter on Trig Integrals.
| If the Function is | then the Antiderivative is | |
|---|---|---|
| Inverse Hyperbolic Sine | \(f(x)=\dfrac{1}{\sqrt{x^2+1}}\) | \(F(x)=\mathrm{arcsinh}\,x+C\) |
| Inverse Hyperbolic Cosine | \(f(x)=\dfrac{1}{\sqrt{x^2-1}}\) | \(F(x)=\mathrm{arccosh}\,x+C\) |
| Inverse Hyperbolic Tangent | \(f(x)=\dfrac{1}{x^2-1}\) | \(F(x)=\mathrm{arctanh}\,x+C\) |
| Inverse Hyperbolic Cotangent\(^\text{3}\) | \(f(x)=\dfrac{1}{x^2-1}\) | \(F(x)=-\,\mathrm{arccoth}\,x+C\) |
| Inverse Hyperbolic Secant | \(f(x)=\dfrac{1}{\sqrt{x^2(1-x^2)}}\) | \(F(x)=-\,\mathrm{arcsech}\,x+C\) |
| Inverse Hyperbolic Cosecant | \(f(x)=\dfrac{1}{\sqrt{x^2(1+x^2)}}\) | \(F(x)=-\,\mathrm{arccsch}\,x+C\) |
\(^\text{2}\) These are the functions whose antiderivatives are the inverse hyperbolic trig functions. We will discuss the antiderivatives of the inverse hyperbolic trig functions in Calculus 2 in the chapter on Integration by Parts.
\(^\text{3}\) This antiderivative formula is not useful, since the formula using \(\mathrm{arctanh}\,x\) is easier. They just differ by a constant.
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