Trigonometric substitution 

Topics in calculus

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

Product rule
Quotient rule
Chain rule
Change of variables
Implicit differentiation
Taylor's theorem
Related rates
List of differentiation identities
Integration

Lists of integrals
Improper integrals
Integration by:
parts, disks, cylindrical
shells, substitution,
trigonometric substitution,
partial fractions, changing order

In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing radical expressions:

\text{For } \sqrt{a^2-x^2} \text{ let } x=a \sin(\theta) \text{ and use } 1-\sin^2(\theta) = \cos^2(\theta)
\text{For } \sqrt{a^2+x^2} \text{ let } x=a \tan(\theta) \text{ and use } 1+\tan^2(\theta) = \sec^2(\theta)
\text{For } \sqrt{x^2-a^2} \text{ let } x=a \sec(\theta) \text{ and use } \sec^2(\theta)-1 = \tan^2(\theta)

Contents

Examples

Integrals containing a2 − x2

In the integral

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

we may use

x=a\sin(\theta),\ dx=a\cos(\theta)\,d\theta
\theta=\arcsin\left(\frac{x}{a}\right)

so that the integral becomes

\int\frac{dx}{\sqrt{a^2-x^2}} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2-a^2\sin^2(\theta)}} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2(1-\sin^2(\theta))}}
{} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}} = \int d\theta=\theta+C=\arcsin\left(\frac{x}{a}\right)+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}.

Some care is needed when picking the bounds. The integration above 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.

Integrals containing a2 + x2

In the integral

\int\frac{dx}{a^2+x^2}

we may write

x=a\tan(\theta),\ dx=a\sec^2(\theta)\,d\theta
\theta=\arctan\left(\frac{x}{a}\right)

so that the integral becomes

\begin{align} & {} \quad \int\frac{dx}{a^2+x^2} = \int\frac{a\sec^2(\theta)\,d\theta}{a^2+a^2\tan^2(\theta)} = \int\frac{a\sec^2(\theta)\,d\theta}{a^2(1+\tan^2(\theta))} \\ & {} = \int \frac{a\sec^2(\theta)\,d\theta}{a^2\sec^2(\theta)} = \int \frac{d\theta}{a} = \frac{\theta}{a}+C = \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C \end{align}

(provided a > 0).

Integrals containing x2 − a2

Integrals like

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

should be done by partial fractions rather than trigonometric substitutions. However, the integral

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

can be done by substitution:

x = a \sec(\theta),\ dx = a \sec(\theta)\tan(\theta)\,d\theta
\theta = \arcsec\left(\frac{x}{a}\right)
\begin{align} & {} \quad \int\sqrt{x^2 - a^2}\,dx = \int\sqrt{a^2 \sec^2(\theta) - a^2} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\ & {} = \int\sqrt{a^2 (\sec^2(\theta) - 1)} \cdot a \sec(\theta)\tan(\theta)\,d\theta = \int\sqrt{a^2 \tan^2(\theta)} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\ & {} = \int a^2 \sec(\theta)\tan^2(\theta)\,d\theta = a^2 \int \sec(\theta)\ (\sec^2(\theta) - 1)\,d\theta \\ & {} = a^2 \int (\sec^3(\theta) - \sec(\theta))\,d\theta. \end{align}

We can then solve this using the formula for the integral of secant cubed.

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

See also