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Trigonometric substitution

From Biocrawler, the free encyclopedia.

Topics in calculus

Differentiation

Integration

In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities

$1-\sin^2\theta\equiv\cos^2\theta$

$1+\tan^2\theta\equiv\sec^2\theta$

$\sec^2\theta-1\equiv\tan^2\theta$

to simplify certain integrals containing the radical expressions

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

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

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

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.

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Examples

In the integral

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

one may use

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

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

a 2 - x 2 = a 2 - a 2 sin 2 (θ) = a 2 (1 - sin 2 (θ)) = a 2 cos 2 (θ),

so that the integral becomes

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

(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

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

In the integral

$\int\frac{1}{a^2+x^2}\,dx$

one may write

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

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

a 2 + x 2 = a 2 + a 2 tan 2 (θ) = a 2 (1 + tan 2 (θ)) = a 2 sec 2 (θ),

x / a = tan(θ),

so that the integral becomes

$\int\frac{1}{a^2\sec^2(\theta)}\,a\sec^2(\theta)\,d\theta =\frac{1}{a}\int\,d\theta=\frac{\theta}{a}+C=\frac{1}{a}\arctan(x/a)+C$

(provided a > 0).

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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)

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

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

$\int\frac{\cos x}{(1+\cos x)^3}\,dx$ $=\int\frac2{1+u^2}\frac{\frac{1-u^2}{1+u^2}}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\,du$ $=\frac14\int(1-u^4)\,du$ $=\frac14\left(u-\frac15u^5\right)+C$ $=\frac{(1+3\cos x+\cos^2x)\sin x}{5(1+\cos x)^3}+C$