Trigonometric & Hyperbolic Substitution

The trigonometric functions ($\sin ,\cos ,\tan$) and their hyperbolic equivalents ($\sinh ,\cosh ,\tanh$) can be used in integrals through the use of their identities. The process can be thought of as a reverse U-Substitution.

Definition §

Just as u-substitution uses the rule: ${\int }_a^{b}f(g(x))g^{\prime }(x)dx={\int }_{g(a)}^{g(b)}f(u)du$ We can ‘reverse’ the rule and perform another change of variables to simplify the integral:

${\int }_a^{b}f(x)dx={\int }_{g^{-1}(a)}^{g^{-1}(b)}f(g(u))g^{\prime }(u)du$ (where $g^{-1}$ is the inverse function of $g$)

The following table helps quickly identify what type of trigonometric/hyperbolic substitution is needed, given a certain integral form.

Integrand	Substitution
$\sqrt{a^2-x^2}$	$x=a\sin (u)$
$\sqrt{a^2+x^2}$	$x=a\sinh (u)$
$\sqrt{x^2-a^2}$	$x=a\cosh (u)$
$\frac{1}{a^2+x^2}$	$x=a\tan (u)$

Examples §

1: Cosh Substitution

Solve the following integral using a trigonometric or hyperbolic substitution: $\int \frac{1}{\sqrt{x^2-25}}dx$
The integral is of the form $\sqrt{x^2-a^2}$. so we’ll be using a cosh substitution: Let $x=g(u)=5\cosh (u)$. Then $g^{\prime }(u)=5\sinh (u)$ The integral can now be expressed in terms of $u$: / $\int \frac{1}{\sqrt{25{\cosh }^2(u)-25}}\times 5\sinh (u)du$ $=\int \frac{5\sinh (u)}{5\sqrt{{\cosh }^2(u)-1}}du$ $=\int \frac{\sinh (u)}{\sinh (u)}du$ $=\int 1du$ $=u+c$ $=\text{arccosh}\left(\frac{x}{5}\right)+c$