# What is trigonometric substitution and why does it work?

Nov 16, 2016

Trig substitution is an integration substitution involving a trig function. It used to solve problem such as

$\int \sqrt{{a}^{2} \pm {x}^{2}} \mathrm{dx}$, and $\int \sqrt{{x}^{2} \pm {1}^{2}} \mathrm{dx}$
$\int \frac{1}{\sqrt{{a}^{2} \pm {x}^{2}}} \mathrm{dx}$, and $\int \frac{1}{\sqrt{{x}^{2} \pm {1}^{2}}} \mathrm{dx}$

and various other similar forms. They work simply because of the various trig identities

Example:

$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Let $x = \sin u \implies \frac{\mathrm{dx}}{\mathrm{du}} = \cos u$,
Hence $\int \ldots \mathrm{dx} = \int . . \cos u \mathrm{du}$

Using the trig identity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have

${\sin}^{2} u + {\cos}^{2} u = 1$
$\therefore {\cos}^{2} u = 1 - {\sin}^{2} u$
$\therefore {\cos}^{2} u = 1 - {x}^{2}$
$\therefore \cos u = \sqrt{1 - {x}^{2}}$

Substituting into the integral we have:

$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\cos} u \cdot \cos \mathrm{du}$
$\therefore \int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \mathrm{du}$
$\therefore \int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = u + C$
$\therefore \int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \arcsin x + C$