# How do you integrate hyperbolic trig functions?

Mar 22, 2015

The easiest way to integrate (or differentiate) the hyperbolic functions is to use their definitions:

$\sinh \left(x\right) = \frac{{e}^{x} - {e}^{- x}}{2}$
$\cosh \left(x\right) = \frac{{e}^{x} + {e}^{- x}}{2}$
$\tanh \left(x\right) = \sinh \frac{x}{\cosh} \left(x\right) = \frac{{e}^{x} - {e}^{- x}}{{e}^{x} + {e}^{- x}}$
$\coth \left(x\right) = \cosh \frac{x}{\sinh} \left(x\right) = \frac{{e}^{x} + {e}^{- x}}{{e}^{x} - {e}^{- x}}$

From here, it should be reasonably straightforward to show that

$\int \sinh \left(x\right) \mathrm{dx} = \cosh \left(x\right) + C$
$\int \cosh \left(x\right) \mathrm{dx} = \sinh \left(x\right) + C$
int tanh(x)dx = ln(cosh((x)) + C
$\int \coth \left(x\right) \mathrm{dx} = \ln \left(\sinh \left(x\right)\right) + C$

where C is the constant of integration. I will show the first two here:

$\int \sinh \left(x\right) \mathrm{dx} = \int \frac{{e}^{x} - {e}^{-} x}{2} = \int {e}^{x} / 2 - {e}^{- x} / 2 \mathrm{dx}$
$= {e}^{x} / 2 - \frac{- {e}^{-} x}{2} + C$ (where $C$ is the constant of integration)
$= {e}^{x} / 2 + {e}^{- x} / 2 + C$
$= \cosh \left(x\right) + C$.

Similarly,
$\int \cosh \left(x\right) \mathrm{dx} = \int {e}^{x} / 2 + {e}^{- x} / 2 \mathrm{dx}$
$= {e}^{x} / 2 + \frac{- {e}^{-} x}{2} + C$
$= {e}^{x} / 2 - {e}^{-} \frac{x}{2} + C$
$= \sinh \left(x\right) + C$.