# 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 radical expressions:[1][2]

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

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

and use the identity

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

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

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

and use the identity

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

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

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

and use the identity

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

## Examples

### Integrals containing a2 − x2

In the integral

${\displaystyle \int {\frac {\sqrt -x^}}},}$

we may use

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

Then,

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

The above step requires that a > 0 and cos(θ) > 0; we can choose a to be the positive square root of a2, and we impose the restriction π/2 < θ < π/2 on θ 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,

${\displaystyle \int _^{\frac {\sqrt -x^}}}=\int _^{\pi /6}d\theta ={\frac {\pi }}.}$

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. Neglecting this restriction, one might have picked θ to go from π to 5π/6, which would have resulted in the negative of the actual value.

### Integrals containing a2 + x2

In the integral

${\displaystyle \int {\frac +x^}}}$

we may write

${\displaystyle x=a\tan \theta ,\quad dx=a\sec ^\theta \,d\theta ,\quad \theta =\arctan {\frac },}$

so that the integral becomes

${\displaystyle {\begin\int {\frac +x^}}&=\int {\frac \theta \,d\theta }+a^\tan ^\theta }}\\&=\int {\frac \theta \,d\theta }(1+\tan ^\theta )}}\\&=\int {\frac \theta \,d\theta }\sec ^\theta }}\\&=\int {\frac }\\&={\frac {\theta }}+C\\&={\frac }\arctan {\frac }+C,\end}}$

provided a ≠ 0.

### Integrals containing x2 − a2

Integrals like

${\displaystyle \int {\frac -a^}}}$

can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral

${\displaystyle \int {\sqrt -a^}}\,dx}$

cannot. In this case, an appropriate substitution is:

${\displaystyle x=a\sec \theta ,\quad dx=a\sec \theta \tan \theta \,d\theta ,\quad \theta =\operatorname {\frac }.}$

Then,

${\displaystyle {\begin\int {\sqrt -a^}}\,dx&=\int {\sqrt \sec ^\theta -a^}}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int {\sqrt (\sec ^\theta -1)}}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int {\sqrt \tan ^\theta }}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int a^\sec \theta \tan ^\theta \,d\theta \\&=a^\int (\sec \theta )(\sec ^\theta -1)\,d\theta \\&=a^\int (\sec ^\theta -\sec \theta )\,d\theta .\end}}$

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,

${\displaystyle {\begin\int f(\sin(x),\cos(x))\,dx&=\int {\frac {\pm {\sqrt }}}}f\left(u,\pm {\sqrt }}\right)\,du&&u=\sin(x)\\\int f(\sin(x),\cos(x))\,dx&=\int {\frac {\mp {\sqrt }}}}f\left(\pm {\sqrt }},u\right)\,du&&u=\cos(x)\\\int f(\sin(x),\cos(x))\,dx&=\int {\frac }}f\left({\frac }},{\frac }}}\right)\,du&&u=\tan \left({\tfrac }\right)\\\int {\frac {\cos x}{(1+\cos x)^}}\,dx&=\int {\frac }}{\frac {\frac }}}{\left(1+{\frac }}}\right)^}}\,du=\int (1-u^)(1+u^)\,du\end}}$

## Hyperbolic substitution

Substitutions of hyperbolic functions can also be used to simplify integrals.[3]

In the integral ${\displaystyle \int {\frac {\sqrt +x^}}}\,dx}$, make the substitution ${\displaystyle x=a\sinh }$, ${\displaystyle dx=a\cosh u\,du.}$

Then, using the identities ${\displaystyle \cosh ^(x)-\sinh ^(x)=1}$ and ${\displaystyle \sinh ^{-1}=\ln(x+{\sqrt +1}}),}$

${\displaystyle {\begin\int {\frac {\sqrt +x^}}}\,dx&=\int {\frac {\sqrt +a^\sinh ^u}}}\,du\\&=\int {\frac }}}}}\,du\\&=\int {\frac }}\,du\\&=u+C\\&=\sinh ^{-1}{\frac }+C\\&=\ln \left({\sqrt {{\frac }}}+1}}+{\frac }\right)+C\\&=\ln \left({\frac {{\sqrt +a^}}+x}}\right)+C\end}}$

## References

1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5.
2. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 0-321-58876-2.
3. ^ Boyadzhiev, Khristo N. "Hyperbolic Substitutions for Integrals" (PDF). Retrieved 4 March 2013.