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

<math>1-\sin^2\theta=\cos^2\theta<math>

<math>1+\tan^2\theta=\sec^2\theta<math>

<math>\sec^2\theta-1=\tan^2\theta<math>

to simplify certain integrals containing the radical expressions

<math>\sqrt{a^2-x^2}<math>

<math>\sqrt{a^2+x^2}<math>

<math>\sqrt{x^2-a^2}<math>

respectively.

In the expression a² − x², the substitution of a sin(θ) for x makes it possible to use the identity 1 − sin²θ = cos²θ.

In the expression a² + x², the substitution of a tan(θ) for x makes it possible to use the identity tan²θ + 1 = sec²θ.

Similarly, in x² − a², the substitution of sec(θ) for x makes it possible to use the identity sec² − 1 = tan².

Examples

In the integral

<math>\int\frac{dx}{\sqrt{a^2-x^2}}<math>

one may use

<math>x=a\sin(\theta)\ \ \mbox{so}\ \mbox{that}\ \sin^{-1}(x/a)=\theta,<math>

<math>dx=a\cos(\theta)\,d\theta,<math>

<math>a^2-x^2=a^2-a^2\sin^2(\theta)=a^2(1-\sin^2(\theta))=a^2\cos^2(\theta),<math>

so that the integral becomes

<math>\int\frac{dx}{\sqrt{a^2-x^2}}=\int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}}

=\int d\theta=\theta+C=\sin^{-1}(x/a)+C<math>

(provided a > 0; if a < 0 then √a² would be |a|, which would differ from a).

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

<math>\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}

=\int_0^{\pi/6}d\theta=\frac{\pi}{6}.<math>

---

In the integral

<math>\int\frac{1}{a^2+x^2}\,dx<math>

one may write

<math>x=a\tan(\theta),\ \mbox{so}\ \mbox{that}\ \theta=\arctan(x/a),<math>

<math>dx=a\sec^2(\theta)\,d\theta,<math>

<math>a^2+x^2=a^2+a^2\tan^2(\theta)=a^2(1+\tan^2(\theta))

=a^2\sec^2(\theta),<math>

<math>x/a=\tan(\theta),<math>

so that the integral becomes

<math>\int\frac{1}{a^2\sec^2(\theta)}\,a\sec^2(\theta)\,d\theta

=\frac{1}{a}\int\,d\theta=\frac{\theta}{a}+C=\frac{1}{a}\arctan(x/a)+C<math>

(provided a > 0).

Substitutions that eliminate trigonometric functions

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

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

<math>\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<math>, <math>u=\cos x<math>

(but be careful with the signs)

<math>\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<math>, <math>u=\tan\frac x2<math>

Example (see quintic of l'Hospital[1] (http://www.mathcurve.com/courbes2d/quintique%20de%20l%27hospital/quintique%20de%20l%27hospital)):

<math>\int\frac{\cos x}{(1+\cos x)^3}\,dx<math><math>

=\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<math><math> =\frac14\int(1-u^4)\,du<math><math> =\frac14\left(u-\frac15u^5\right)+C<math><math> =\frac{(1+3\cos x+\cos^2x)\sin x}{5(1+\cos x)^3}+C<math>

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