# Euler substitution

Euler substitution is a method for evaluating integrals of the form:

$\int\!R(x,\sqrt{ax^2+bx+c})\, \mathrm dx$

where $R$ is a rational function of $x$ and $\sqrt{ax^2+bx+c}$. In such cases, the integrand can be changed to a rational function by using the substitutions of Euler.[1]

## The first substitution of Euler

The first substitution of Euler is used when $a > 0$ . We substitute $\sqrt{ax^2+bx+c} = \pm x\sqrt{a}+t$ and solve the resulting expression for $x$. We have that $x = \frac{c-t^2}{\pm 2t\sqrt{a}-b}$ and that the $\mathrm dx$ term is expressible rationally via $t$.

In this substitution, either the positive sign or the negative sign can be chosen.

## The second substitution of Euler

If $c > 0$ we take \begin{align} \sqrt{ax^2+bx+c} = xt\pm\sqrt{c}. \end{align} We solve for $x$ similarly as above and find, $x = \frac{\pm 2t\sqrt{c}-b}{a-t^2}.$

Again, either the positive or the negative sign can be chosen.

## The third substitution of Euler

If the polynomial $ax^2+bx+c$ has real roots $\alpha$ and $\beta$ we may chose $\sqrt{ax^2 + bx + c} = \sqrt{a(x-\alpha)(x-\beta)} = (x-\alpha)t$. This yields $x = \frac{a\beta-\alpha t^2}{a-t^2}$ and as in the preceding cases, we can express the entire integrand rationally via $t$.

## Examples

In the integral $\int\! \frac{\mathrm dx}{\sqrt{x^2+c}}$ we can use the first substitution and set $\sqrt{x^2+c} = -x+t$, thus

$x = \frac{t^2-c}{2t} \quad\quad \mathrm dx = \frac{t^2+c}{2t^2}\,\mathrm dt$
$\sqrt{x^2+c} = -\frac{t^2-c}{2t}+t = \frac{t^2+c}{2t}$

Accordingly we obtain:

$\int \frac{\mathrm dx}{\sqrt{x^2+c}} = \int \frac{\frac{t^2+c}{2t^2}}{\frac{t^2+c}{2t}}\, \mathrm dt$
$= \int\!\frac{\mathrm dt}{t} = \ln|t|+C = \ln|x+\sqrt{x^2+c}|+C$

The cases $c = \pm 1$, give the formulas

$\int \frac{\mathrm dx}{\sqrt{x^2+1}} = \mbox{arsinh}(x) + C$
$\int \frac{\mathrm dx}{\sqrt{x^2-1}} = \mbox{arcosh}(x) + C \quad (x > 1)$

## Generalizations

The substitutions of Euler can be generalized by allowing the use of imaginary numbers. For example, in the integral $\textstyle \int\! \frac{\mathrm dx}{\sqrt{-x^2+c}}$, the substitution $\sqrt{x^2+c} = \pm ix+t$ can be used. Extensions to the complex numbers allows us to use every type of Euler substitution regardless of the coefficients on the quadratic.

The substitutions of Euler can be generalized to a larger class of functions. Consider integrals of the form

$\int\!R_1(x,\sqrt{ax^2+bx+c})\,\log(R_2(x,\sqrt{ax^2+bx+c}))\, \mathrm dx$

where $R_1$ and $R_2$ are rational functions of $x$ and $\sqrt{ax^2+bx+c}$. This integral can be transformed by the substitution $\sqrt{ax^2+bx+c} = \sqrt{a}+xt$ into another integral

$\int\!\overset{\sim}{R_1}(t)\,\log(\overset{\sim}{R_2}(t))\, \mathrm dt$

where $\overset{\sim}{R_1}(t)$ and $\overset{\sim}{R_2}(t)$ are now simply rational functions of $t$. In principle, factorization and partial fraction decomposition can be employed to break the integral down into simple terms which can be integrated analytically through use of the dilogarithm function.[2]

## References

1. ^ N. Piskunov, Diferentsiaal- ja integraalarvutus körgematele tehnilistele öppeasutustele. Viies, taiendatud trukk. Kirjastus Valgus, Tallinn (1965). Note: Euler substitutions can be found in most Russian calculus textbooks.
2. ^ Zwillinger, Daniel. The Handbook of Integration. 1992: Jones and Bartlett. pp. 145–146. ISBN 978-0867202939.