Refresher - Trigonometric Substitution

Trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing either

$a^{2} - u^{2}, a^{2} + u^{2}, or u^{2} - a^{2}$ where $a$ is a constant and $u$ is a variable.

Whenever an integral contains $a^{2} - u^{2}$ , substitute $u = a sin θ, d u = a cos θ d θ$ , and then use the identity $1 - {sin}^{2} θ = {cos}^{2} θ .$

Example:

In the integral

$LaTeX: \int\frac{dx}{\sqrt{a^2-x^2}}$ $\int\frac{dx}{\sqrt{a^2-x^2}}$

we may use

$x = a sin θ, d x = a cos θ d θ$

$LaTeX: \theta =\arcsin(\frac{x}{a})$ $\theta =\arcsin(\frac{x}{a})$

so that the integral becomes

$$\int\frac{dx}{\sqrt{a^2-x^2}}=\int\frac{a\cos\theta d\theta}{\sqrt{a^2-a^2\sin^2\theta}}=\int\frac{a\cos\theta d\theta}{\sqrt{a^2(1-\sin^2\theta)}}=\int\frac{a\cos\theta d\theta}{\sqrt{a^2\cos^2\theta}}=\int d\theta=\theta+C=\arcsin(\frac{x}{a})+C$.$

Note that the above step requires that $a > 0$ and $cos θ > 0$ ; we can choose the $a$ to be the positive square root of $a^{2}$ ; and we impose the restriction on $θ$ to be $- \frac{π}{2} < θ < \frac{π}{2}$ by using the arcsine function.

Whenever an integral contains $u^{2} - a^{2}$ , substitute $u = a sec θ, d u = a sec θ tan θ d θ$ , and then use the identity ${sec}^{2} θ - 1 = {tan}^{2} θ$ .

Example

Integrals like

$LaTeX: \int\frac{dx}{x^2-a^2}$ $\int\frac{dx}{x^2-a^2}$

should be done by partial fractions rather than trigonometric substitutions. However, the integral

$LaTeX: \int\sqrt{x^2-a^2}dx$ $\int\sqrt{x^2-a^2}dx$

can be done by trigonometric substitution:

$x = a sec θ, d x = a sec θ tan θ d θ$

$LaTeX: \theta=\sec^{-1}(\frac{x}{a})$ $\theta=\sec^{-1}(\frac{x}{a})$

$LaTeX: \int\sqrt{x^2-a^2}dx=\int\sqrt{a^2\sec^2\theta-a^2}sec\theta \tan\theta d\theta=\int\sqrt{a^2(\sec^2\theta-1)}a\sec\theta \tan\theta d\theta=\int\sqrt{a^2\tan^2\theta}a\sec\theta\tan\theta d\theta=\int a^2\sec\theta\tan^2\theta d\theta=a^2\int\sec\theta(1-\sec^2\theta)d\theta=a^2\int(\sec\theta-\sec^3\theta)d\theta$ $\int\sqrt{x^2-a^2}dx=\int\sqrt{a^2\sec^2\theta-a^2}sec\theta \tan\theta d\theta=\int\sqrt{a^2(\sec^2\theta-1)}a\sec\theta \tan\theta d\theta=\int\sqrt{a^2\tan^2\theta}a\sec\theta\tan\theta d\theta=\int a^2\sec\theta\tan^2\theta d\theta=a^2\int\sec\theta(1-\sec^2\theta)d\theta=a^2\int(\sec\theta-\sec^3\theta)d\theta$