# Trigonometric substitution

 Topics in calculus Differentiation Integration Integration by substitution | Integration by parts | Integration by trigonometric substitution | Solids of revolution | Integration by disks | Integration by cylindrical shells | Improper integrals | Lists of integrals

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

[itex]1-\sin^2\theta\equiv\cos^2\theta[itex]
[itex]1+\tan^2\theta\equiv\sec^2\theta[itex]
[itex]\sec^2\theta-1\equiv\tan^2\theta[itex]

to simplify certain integrals containing the radical expressions

[itex]\sqrt{a^2-x^2}[itex]
[itex]\sqrt{a^2+x^2}[itex]
[itex]\sqrt{x^2-a^2}[itex]

respectively.

In the expression a2x2, 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 x2a2, the substitution of sec(θ) for x makes it possible to use the identity sec2 − 1 = tan2.

## Examples

In the integral

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

one may use

[itex]x=a\sin(\theta)\ \ \mbox{so}\ \mbox{that}\ \sin^{-1}(x/a)=\theta,[itex]
[itex]dx=a\cos(\theta)\,d\theta,[itex]
[itex]a^2-x^2=a^2-a^2\sin^2(\theta)=a^2(1-\sin^2(\theta))=a^2\cos^2(\theta),[itex]

so that the integral becomes

[itex]\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[itex]

(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

[itex]\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}

=\int_0^{\pi/6}d\theta=\frac{\pi}{6}.[itex]

In the integral

[itex]\int\frac{1}{a^2+x^2}\,dx[itex]

one may write

[itex]x=a\tan(\theta),\ \mbox{so}\ \mbox{that}\ \theta=\arctan(x/a),[itex]
[itex]dx=a\sec^2(\theta)\,d\theta,[itex]
[itex]a^2+x^2=a^2+a^2\tan^2(\theta)=a^2(1+\tan^2(\theta))

=a^2\sec^2(\theta),[itex]

[itex]x/a=\tan(\theta),[itex]

so that the integral becomes

[itex]\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[itex]

(provided a > 0).

## Substitutions that eliminate trigonometric functions

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

(but be careful with the signs)

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

[itex]\int\frac{\cos x}{(1+\cos x)^3}\,dx[itex][itex]

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

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