# Trigonometric substitution

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Template:Wikiversity Template:Sister {{#invoke: Sidebar | collapsible }} 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:[1][2]

Substitution 1. If the integrand contains a2 − x2, let

${\displaystyle x=a\sin(\theta )}$

and use the identity

${\displaystyle 1-\sin ^{2}(\theta )=\cos ^{2}(\theta ).}$

Substitution 2. If the integrand contains a2 + x2, let

${\displaystyle x=a\tan(\theta )}$

and use the identity

${\displaystyle 1+\tan ^{2}(\theta )=\sec ^{2}(\theta ).}$

Substitution 3. If the integrand contains x2 − a2, let

${\displaystyle x=a\sec(\theta )}$

and use the identity

${\displaystyle \sec ^{2}(\theta )-1=\tan ^{2}(\theta ).}$

## Examples

### Integrals containing a2 − x2

In the integral

${\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {a^{2}-x^{2}}}}}$

we may use

${\displaystyle x=a\sin(\theta ),\quad \mathrm {d} x=a\cos(\theta )\,\mathrm {d} \theta ,\quad \theta =\arcsin \left({\frac {x}{a}}\right)}$
{\displaystyle {\begin{aligned}\int {\frac {\mathrm {d} x}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos(\theta )\,\mathrm {d} \theta }{\sqrt {a^{2}-a^{2}\sin ^{2}(\theta )}}}\\&=\int {\frac {a\cos(\theta )\,\mathrm {d} \theta }{\sqrt {a^{2}(1-\sin ^{2}(\theta ))}}}\\&=\int {\frac {a\cos(\theta )\,\mathrm {d} \theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}\\&=\int \mathrm {d} \theta \\&=\theta +C\\&=\arcsin \left({\tfrac {x}{a}}\right)+C\end{aligned}}}

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

${\displaystyle \int _{0}^{\frac {a}{2}}{\frac {\mathrm {d} x}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\frac {\pi }{6}}\mathrm {d} \theta ={\tfrac {\pi }{6}}.}$

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 give us the negative of the result.

### Integrals containing a2 + x2

In the integral

${\displaystyle \int {\frac {\mathrm {d} x}{a^{2}+x^{2}}}}$

we may write

${\displaystyle x=a\tan(\theta ),\quad \mathrm {d} x=a\sec ^{2}(\theta )\,\mathrm {d} \theta ,\quad \theta =\arctan \left({\tfrac {x}{a}}\right)}$

so that the integral becomes

{\displaystyle {\begin{aligned}\int {\frac {\mathrm {d} x}{a^{2}+x^{2}}}&=\int {\frac {a\sec ^{2}(\theta )\,\mathrm {d} \theta }{a^{2}+a^{2}\tan ^{2}(\theta )}}\\&=\int {\frac {a\sec ^{2}(\theta )\,\mathrm {d} \theta }{a^{2}(1+\tan ^{2}(\theta ))}}\\&=\int {\frac {a\sec ^{2}(\theta )\,\mathrm {d} \theta }{a^{2}\sec ^{2}(\theta )}}\\&=\int {\frac {\mathrm {d} \theta }{a}}\\&={\tfrac {\theta }{a}}+C\\&={\tfrac {1}{a}}\arctan \left({\tfrac {x}{a}}\right)+C\end{aligned}}}

(provided a ≠ 0).

### Integrals containing x2 − a2

Integrals like

${\displaystyle \int {\frac {\mathrm {d} x}{x^{2}-a^{2}}}}$

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

${\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,\mathrm {d} x}$

can be done by substitution:

${\displaystyle x=a\sec(\theta ),\quad \mathrm {d} x=a\sec(\theta )\tan(\theta )\,\mathrm {d} \theta ,\quad \theta =\operatorname {arcsec} \left({\tfrac {x}{a}}\right)}$
{\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,\mathrm {d} x&=\int {\sqrt {a^{2}\sec ^{2}(\theta )-a^{2}}}\cdot a\sec(\theta )\tan(\theta )\,\mathrm {d} \theta \\&=\int {\sqrt {a^{2}(\sec ^{2}(a)-1)}}\cdot a\sec(\theta )\tan(\theta )\,\mathrm {d} \theta \\&=\int {\sqrt {a^{2}\tan ^{2}(\theta )}}\cdot a\sec(\theta )\tan(\theta )\,\mathrm {d} \theta \\&=\int a^{2}\sec(\theta )\tan ^{2}(\theta )\,\mathrm {d} \theta \\&=a^{2}\int \sec(\theta )(\sec ^{2}(\theta )-1)\,\mathrm {d} \theta \\&=a^{2}\int (\sec ^{3}(\theta )-\sec(\theta ))\,\mathrm {d} \theta .\end{aligned}}}

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. In particular, see Tangent half-angle substitution.

For instance,

${\displaystyle \int {\frac {1}{\sqrt {a^{2}+x^{2}}}}\,\mathrm {d} x}$