Euler substitution

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Euler substitution is a method for evaluating integrals of the form

where ${\displaystyle R}$ is a rational function of ${\displaystyle x}$ and ${\textstyle {\sqrt {ax^{2}+bx+c}}}$. 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 ${\displaystyle a>0}$. We substitute

and solve the resulting expression for ${\displaystyle x}$. We have that ${\displaystyle x={\frac {c-t^{2}}{\pm 2t{\sqrt {a}}-b}}}$ and that the ${\displaystyle dx}$ term is expressible rationally in ${\displaystyle t}$.

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

Euler's second substitution

If ${\displaystyle c>0}$, we take

We solve for ${\displaystyle x}$ similarly as above and find

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

Euler's third substitution

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

Worked examples

Examples for Euler's first substitution

One

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

Accordingly, we obtain:

The cases ${\displaystyle c=\pm 1}$ give the formulas

Two

For finding the value of

we find ${\displaystyle t}$ using the first substitution of Euler, ${\textstyle {\sqrt {x^{2}+4x-4}}={\sqrt {1}}x+t=x+t}$. Squaring both sides of the equation gives us ${\displaystyle x^{2}+4x-4=x^{2}+2xt+t^{2}}$, from which the ${\displaystyle x^{2}}$ terms will cancel out. Solving for ${\displaystyle x}$ yields

From there, we find that the differentials ${\displaystyle dx}$ and ${\displaystyle dt}$ are related by ${\displaystyle dx={\frac {-2t^{2}+8t+8}{(4-2t)^{2}}}dt.}$

Hence,

Examples for Euler's second substitution

In the integral

we can use the second substitution and set ${\displaystyle {\sqrt {-x^{2}+x+2}}=xt+{\sqrt {2}}}$. Thus and

Accordingly, we obtain:

Examples for Euler's third substitution

To evaluate

we can use the third substitution and set ${\textstyle {\sqrt {-(x-2)(x-1)}}=(x-2)t}$. Thus and

Next,

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 ${\textstyle \int {\frac {dx}{\sqrt {-x^{2}+c}}}}$, the substitution ${\textstyle {\sqrt {-x^{2}+c}}=\pm ix+t}$ 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

where ${\displaystyle R_{1}}$ and ${\displaystyle R_{2}}$ are rational functions of ${\displaystyle x}$ and ${\textstyle {\sqrt {ax^{2}+bx+c}}}$. This integral can be transformed by the substitution ${\textstyle {\sqrt {ax^{2}+bx+c}}={\sqrt {a}}+xt}$ into another integral where ${\displaystyle {\tilde {R}}_{1}(t)}$ and ${\displaystyle {\tilde {R}}_{2}(t)}$ are now simply rational functions of ${\displaystyle t}$. 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

References

This article incorporates material from Eulers Substitutions For Integration on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.