Euler substitution explained

Euler substitution is a method for evaluating integrals of the form

$$\backslash int\; R(x,\; \backslash sqrt)\; \backslash ,\; dx,$$

where

R is a rational function of x and $\backslash sqrt$. In such cases, the integrand can be changed to a rational function by using the substitutions of Euler.^[1]

Euler's first substitution

The first substitution of Euler is used when

a > 0 . We substitute$$\backslash sqrt\; =\; \backslash pm\; x\backslash sqrt\; +\; t$$and solve the resulting expression for x . We have that x = (c - t ^{2})/(\pm 2 t\sqrt{a) - b} and that the dx term is expressible rationally in t .

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

Euler's second substitution

If

c > 0 , we take$$\backslash sqrt\; =\; xt\; \backslash pm\; \backslash sqrt.$$We solve for x similarly as above and find$$x\; =\; \backslash frac.$$

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

Euler's third substitution

If the polynomial

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

Worked examples

Examples for Euler's first substitution

One

In the integral

\int (dx)/(\sqrt{x ^{2+c}) } we can use the first substitution and set $\backslash sqrt\; =\; -x+t$, thus$$x\; =\; \backslash frac\; \backslash quad\backslash quad\; \backslash \; dx\; =\; \backslash frac\backslash ,\backslash \; dt$$$$\backslash sqrt\; =\; -\backslash frac+t\; =\; \backslash frac$$Accordingly, we obtain:$$\backslash int\; \backslash frac=\; \backslash int\; \backslash frac\backslash ,\; \backslash \; dt=\; \backslash int\; \backslash frac=\; \backslash ln|t|+C=\; \backslash ln\backslash left|x+\backslash sqrt\backslash right|+C$$

The cases

c = \pm 1 give the formulas

Two

For finding the value of $$\backslash int\backslash fracdx,$$ we find

t using the first substitution of Euler, $\backslash sqrt\; =\; \backslash sqrtx+t\; =\; x+t$. Squaring both sides of the equation gives us x ^{2} +4 x -4 = x ^{2} + 2 xt + t ^{2} , from which the x ^{2} terms will cancel out. Solving for x yields $$x=\backslash frac.$$

From there, we find that the differentials

dx and dt are related bydx = (-2 t ^{2} +8 t +8)/(( 4-2 t) ^{2}) dt.

Hence,

Examples for Euler's second substitution

In the integral $$\backslash int\backslash !\; \backslash frac,$$ we can use the second substitution and set

\sqrt{ - x ^{2+x+2}} = xt + \sqrt{ 2 } . Thus$$x\; =\; \backslash frac\; \backslash qquad\; dx\; =\; \backslash frac\; dt,$$and$$\backslash sqrt\; =\; \backslash fract\; +\; \backslash sqrt\; =\; \backslash frac$$

Accordingly, we obtain:

Examples for Euler's third substitution

To evaluate$$\backslash int\backslash !\; \backslash frac\backslash \; dx,$$ we can use the third substitution and set $\backslash sqrt\; =\; (x-2)t$. Thus$$x\; =\; \backslash frac\; \backslash qquad\; \backslash \; dx\; =\; \backslash frac\backslash ,\backslash \; dt,$$and$$\backslash sqrt\; =\; (x-2)t\; =\; \backslash frac$$

Next,$$\backslash int\; \backslash frac\backslash \; dx\; =\; \backslash int\backslash frac\backslash \; dt\; =\; \backslash int\backslash frac\backslash \; dt.$$As we can see this is a rational function which can be solved using partial fractions.

Generalizations

The substitutions of Euler can be generalized by allowing the use of imaginary numbers. For example, in the integral $\backslash int\; \backslash frac$, the substitution $\backslash sqrt\; =\; \backslash pm\; ix\; +\; t$ 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$$\backslash int\; R\_1\; \backslash left(x,\; \backslash sqrt\; \backslash right)\; \backslash ,\; \backslash log\backslash left(R\_2\backslash left(x,\; \backslash sqrt\backslash right)\backslash right)\; \backslash ,\; dx,$$where

R _{1} and R _{2} are rational functions of x and $\backslash sqrt$. This integral can be transformed by the substitution $\backslash sqrt\; =\; \backslash sqrt\; +\; xt$ into another integral$$\backslash int\; \backslash tilde\; R\_1(t)\; \backslash log\backslash big(\backslash tilde\; R\_2(t)\backslash big)\; \backslash ,\; dt,$$where \tilde R _{1(t)} and \tilde 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]

See also

• Integration by substitution
• Trigonometric substitution
• Weierstrass substitution

Notes and 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. Book: Zwillinger . Daniel . The Handbook of Integration . Jones and Bartlett . 978-0867202939 . 145–146.