Trigonometric substitution

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Topics in calculus

Fundamental theorem
Limits of functions
Continuity
Vector calculus
Tensor calculus
Mean value theorem
Differentiation

Product rule
Quotient rule
Chain rule
Implicit differentiation
Taylor's theorem
Related rates
Table of derivatives
Integration

Lists of integrals
Improper integrals
Integration by: parts, disks,
cylindrical shells, substitution,
trigonometric substitution

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.

Contents

[edit] Examples

[edit] Integrals containing a2 − x2

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,
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=\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.)

[edit] Integrals containing a2 + x2

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).

[edit] Integrals containing x2 − a2

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.

[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

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

[edit] See also

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