Trigonometric substitution

Topics in Calculus

Fundamental theorem | Function | Limits of functions | Continuity | Calculus with polynomials

Product rule | Quotient rule | Chain rule | Implicit differentiation | Taylor's theorem

Vector Calculus

Tensor Calculus

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

1 - sin²θ = cos²θ

1 + tan²θ = sec²θ

sec²θ - 1 = tan²θ

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.

Examples

In the integral

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

one may use

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

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

a² - x² = a² - a²sin²(θ) = a²(1 - sin²(θ)) = a²cos²(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

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

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

a² + x² = a² + a²tan²(θ) = a²(1 + tan²(θ)) = a²sec²(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).

Substitutions that eliminate trigonometric functions

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

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

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

(but be careful with the signs)

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

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

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

Categories: Calculus

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