Trigonometric substitution

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In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing radical expressions:

If the integrand contains ${a^2-x^2},\,$

let

$x = a \sin \theta\,$

and use the identity

$1-\sin^2 \theta = \cos^2 \theta.\,$

If the integrand contains

$\sqrt{a^2+x^2}\,$

let $x = a \tan \theta\,$

and use the identity

$1+\tan^2 \theta = \sec^2 \theta.\,$

If the integrand contains

$\sqrt{x^2-a^2}\,$

let

$x = a \sec \theta\,$

and use the identity

$\sec^2 \theta-1 = \tan^2 \theta.\,$

[edit] Examples

[edit] Integrals containing a² − x²

In the integral

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

we may use

$x=a\sin(\theta),\ dx=a\cos(\theta)\,d\theta$

$\theta=\arcsin\left(\frac{x}{a}\right)$

so that the integral becomes

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

${} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}} = \int d\theta=\theta+C=\arcsin\left(\frac{x}{a}\right)+C$

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 a²; and we impose the restriction on θ to be −π/2 < θ < π/2 by using the arcsin function.

For a definite integral, one must figure out how the bounds of integration change. For example, as x goes from 0 to a/2, then sin(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have

$\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}} =\int_0^{\pi/6}d\theta=\frac{\pi}{6}.$

Some care is needed when picking the bounds. The integration above requires that −π/2 < θ < π/2, so θ going from 0 to π/6 is the only choice. If we had missed this restriction, we might have picked θ to go from π to 5π/6, which would result in the negative of the result.

[edit] Integrals containing a² + x²

In the integral

$\int\frac{dx}{a^2+x^2}$

we may write

$x=a\tan(\theta),\ dx=a\sec^2(\theta)\,d\theta$

$\theta=\arctan\left(\frac{x}{a}\right)$

so that the integral becomes

$\begin{align} & {} \quad \int\frac{dx}{a^2+x^2} = \int\frac{a\sec^2(\theta)\,d\theta}{a^2+a^2\tan^2(\theta)} = \int\frac{a\sec^2(\theta)\,d\theta}{a^2(1+\tan^2(\theta))} \\ & {} = \int \frac{a\sec^2(\theta)\,d\theta}{a^2\sec^2(\theta)} = \int \frac{d\theta}{a} = \frac{\theta}{a}+C = \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C \end{align}$

(provided a > 0).

[edit] Integrals containing x² − a²

Integrals like

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

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

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

can be done by substitution:

$x = a \sec(\theta),\ dx = a \sec(\theta)\tan(\theta)\,d\theta$

$\theta = \arcsec\left(\frac{x}{a}\right)$

$\begin{align} & {} \quad \int\sqrt{x^2 - a^2}\,dx = \int\sqrt{a^2 \sec^2(\theta) - a^2} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\ & {} = \int\sqrt{a^2 (\sec^2(\theta) - 1)} \cdot a \sec(\theta)\tan(\theta)\,d\theta = \int\sqrt{a^2 \tan^2(\theta)} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\ & {} = \int a^2 \sec(\theta)\tan^2(\theta)\,d\theta = a^2 \int \sec(\theta)\ (\sec^2(\theta) - 1)\,d\theta \\ & {} = a^2 \int (\sec^3(\theta) - \sec(\theta))\,d\theta. \end{align}$

We can then solve this using the formula for the integral of secant cubed.

[edit] Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions. For instance,

$\int f(\sin x,\cos x)\,dx=\int\frac1{\pm\sqrt{1-u^2}}f\left(u,\pm\sqrt{1-u^2}\right)\,du, \qquad \qquad u=\sin x$

$\int f(\sin x,\cos x)\,dx=\int\frac{-1}{\pm\sqrt{1-u^2}}f\left(\pm\sqrt{1-u^2},u\right)\,du \qquad \qquad u=\cos x$

(but be careful with the signs)

$\int f(\sin x,\cos x)\,dx=\int\frac2{1+u^2} f\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\,du \qquad \qquad u=\tan\frac x2$

$\int\frac{\cos x}{(1+\cos x)^3}\,dx = \int\frac2{1+u^2}\frac{\frac{1-u^2}{1+u^2}}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\,du =$

$\frac{1}{4}\int(1-u^4)\,du = \frac{1}{4}\left(u-\frac15u^5\right) + C = \frac{(1+3\cos x+\cos^2x)\sin x}{5(1+\cos x)^3} + C$

[edit] See also

Tangent half-angle formula

Trigonometric substitution

From Wikipedia, the free encyclopedia

Contents

[edit] Examples

[edit] Integrals containing a² − x²

[edit] Integrals containing a² + x²

[edit] Integrals containing x² − a²

[edit] Substitutions that eliminate trigonometric functions

[edit] See also

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Trigonometric substitution

From Wikipedia, the free encyclopedia

Contents

[edit] Examples

[edit] Integrals containing a2 − x2

[edit] Integrals containing a2 + x2

[edit] Integrals containing x2 − a2

[edit] Substitutions that eliminate trigonometric functions

[edit] See also

Views

Personal tools

Navigation

Search

Interaction

Toolbox

Languages

[edit] Integrals containing a² − x²

[edit] Integrals containing a² + x²

[edit] Integrals containing x² − a²