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 the radical expressions:

$1-\sin^2\theta\;=\;\cos^2\theta$ for $\sqrt{a^2-x^2}$

$1+\tan^2\theta\;=\;\sec^2\theta$ for $\sqrt{a^2+x^2}$

$\sec^2\theta-1\;=\;\tan^2\theta$ to simplify $\sqrt{x^2-a^2}$

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 a sec(θ) for x makes it possible to use the identity sec²θ − 1 = tan²θ.

1 Examples
2 Substitutions that eliminate trigonometric functions
3 See also

In the integral

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

one may use

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

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

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

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

(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 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

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

(Be careful when picking the bounds. The integration from the above section 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.)

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).

Integrals like

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

should be done by partial fractions rather than trigonometric subtstitutions.

The integral

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

can be done by the substitution

$\begin{align} x &{}= a \sec\theta, \\ dx &{}= a \sec\theta\tan\theta\,d\theta, \\ a^2 - x^2 &{}= a^2 \tan^2\theta. \end{align}$

This will involve the integral of secant cubed.

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'Hôspital[1]):

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

tangent half-angle formula

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Category: Integral calculus

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