How to integrate hyperbolic functions

Hyperbolic functions have their own list of antiderivatives, and the signs are easier to follow when we connect each rule to a derivative. We’ll use the list with an inside factor, then integrate an expression with three terms.

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Read the standard hyperbolic antiderivatives

Hyperbolic antiderivatives with input u

Integrand	Antiderivative, plus C
$\sinh u$	$\cosh u$
$\cosh u$	$\sinh u$
$\tanh u$	$\ln \left(\cosh u\right)$
$\coth u$	$\ln \left|\sinh u\right|$
$\mathrm{sech}u$	$\arctan \left(\sinh u\right)$
$\mathrm{csch}u$	$\ln \left|\tanh \right(\frac{u}{2}\left)\right|$
${\mathrm{sech}}^{2}u$	$\tanh u$
${\mathrm{csch}}^{2}u$	$-\coth u$
$\mathrm{sech}u\tanh u$	$-\mathrm{sech}u$
$\mathrm{csch}u\coth u$	$-\mathrm{csch}u$

When $u=ax$ with $a\ne 0$, the substitution gives $dx=du/a$. Keep the antiderivative on an interval where the integrand is defined.

Integrate hyperbolic secant

$\int 5\mathrm{sech}\left(3x\right)dx=\frac{5}{3}\arctan \left(\sinh \right(3x\left)\right)+C$

Integrate a sum one term at a time

$\int (16x^3-2\sinh (3x)+4{\mathrm{csch}}^2(6x\left)\right)dx$

$=4x^4-\frac{2}{3}\cosh \left(3x\right)-\frac{2}{3}\coth \left(6x\right)+C$

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