Euler substitution

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Euler substitution is a method for evaluating integrals of the form

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

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]

The first substitution of Euler is used when ${\displaystyle a>0}$. We substitute $${\displaystyle {\sqrt {ax^{2}+bx+c}}=\pm x{\sqrt {a}}+t}$$ 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.

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

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

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 ${\displaystyle t}$.

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 $${\displaystyle x={\frac {t^{2}-c}{2t}}\quad \quad \ dx={\frac {t^{2}+c}{2t^{2}}}\,\ dt}$$ $${\displaystyle {\sqrt {x^{2}+c}}=-{\frac {t^{2}-c}{2t}}+t={\frac {t^{2}+c}{2t}}}$$ Accordingly, we obtain: $${\displaystyle \int {\frac {dx}{\sqrt {x^{2}+c}}}=\int {\frac {\frac {t^{2}+c}{2t^{2}}}{\frac {t^{2}+c}{2t}}}\,\ dt=\int {\frac {dt}{t}}=\ln |t|+C=\ln \left|x+{\sqrt {x^{2}+c}}\right|+C}$$

The cases ${\displaystyle c=\pm 1}$ give the formulas $${\displaystyle {\begin{aligned}\int {\frac {\ dx}{\sqrt {x^{2}+1}}}&=\operatorname {arsinh} (x)+C\\[6pt]\int {\frac {\ dx}{\sqrt {x^{2}-1}}}&=\operatorname {arcosh} (x)+C\qquad (x>1)\end{aligned}}}$$

For finding the value of $${\displaystyle \int {\frac {1}{x{\sqrt {x^{2}+4x-4}}}}dx,}$$ 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 $${\displaystyle x={\frac {t^{2}+4}{4-2t}}.}$$

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, $${\displaystyle {\begin{aligned}\int {\frac {dx}{x{\sqrt {x^{2}+4x-4}}}}&=\int {\frac {\frac {-2t^{2}+8t+8}{(4-2t)^{2}}}{\left({\frac {t^{2}+4}{4-2t}}\right)\left({\frac {-t^{2}+4t+4}{4-2t}}\right)}}dt&&t={\sqrt {x^{2}+4x-4}}-x\\[6pt]&=2\int {\frac {dt}{t^{2}+4}}=\tan ^{-1}\left({\frac {t}{2}}\right)+C\\[6pt]&=\tan ^{-1}\left({\frac {{\sqrt {x^{2}+4x-4}}-x}{2}}\right)+C\end{aligned}}}$$

In the integral $${\displaystyle \int \!{\frac {dx}{x{\sqrt {-x^{2}+x+2}}}},}$$ we can use the second substitution and set ${\displaystyle {\sqrt {-x^{2}+x+2}}=xt+{\sqrt {2}}}$. Thus $${\displaystyle x={\frac {1-2{\sqrt {2}}t}{t^{2}+1}}\qquad dx={\frac {2{\sqrt {2}}t^{2}-2t-2{\sqrt {2}}}{(t^{2}+1)^{2}}}dt,}$$ and $${\displaystyle {\sqrt {-x^{2}+x+2}}={\frac {1-2{\sqrt {2}}t}{t^{2}+1}}t+{\sqrt {2}}={\frac {-{\sqrt {2}}t^{2}+t+{\sqrt {2}}}{t^{2}+1}}}$$

Accordingly, we obtain: $${\displaystyle {\begin{aligned}\int {\frac {dx}{x{\sqrt {-x^{2}+x+2}}}}&=\int {\frac {\frac {2{\sqrt {2}}t^{2}-2t-2{\sqrt {2}}}{(t^{2}+1)^{2}}}{{\frac {1-2{\sqrt {2}}t}{t^{2}+1}}{\frac {-{\sqrt {2}}t^{2}+t+{\sqrt {2}}}{t^{2}+1}}}}dt\\[6pt]&=\int \!{\frac {-2}{-2{\sqrt {2}}t+1}}dt={\frac {1}{\sqrt {2}}}\int {\frac {-2{\sqrt {2}}}{-2{\sqrt {2}}t+1}}dt\\[6pt]&={\frac {1}{\sqrt {2}}}\ln \left|2{\sqrt {2}}t-1\right|+C\\[4pt]&={\frac {\sqrt {2}}{2}}\ln \left|2{\sqrt {2}}{\frac {{\sqrt {-x^{2}+x+2}}-{\sqrt {2}}}{x}}-1\right|+C\end{aligned}}}$$

To evaluate $${\displaystyle \int \!{\frac {x^{2}}{\sqrt {-x^{2}+3x-2}}}\ dx,}$$ we can use the third substitution and set ${\textstyle {\sqrt {-(x-2)(x-1)}}=(x-2)t}$. Thus $${\displaystyle x={\frac {-2t^{2}-1}{-t^{2}-1}}\qquad \ dx={\frac {2t}{(-t^{2}-1)^{2}}}\,\ dt,}$$ and $${\displaystyle {\sqrt {-x^{2}+3x-2}}=(x-2)t={\frac {t}{-t^{2}-1.}}}$$

Next, $${\displaystyle \int {\frac {x^{2}}{\sqrt {-x^{2}+3x-2}}}\ dx=\int {\frac {\left({\frac {-2t^{2}-1}{-t^{2}-1}}\right)^{2}{\frac {2t}{(-t^{2}-1)^{2}}}}{\frac {t}{-t^{2}-1}}}\ dt=\int {\frac {2(-2t^{2}-1)^{2}}{(-t^{2}-1)^{3}}}\ dt.}$$ As we can see this is a rational function which can be solved using partial fractions.

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 $${\displaystyle \int R_{1}\left(x,{\sqrt {ax^{2}+bx+c}}\right)\,\log \left(R_{2}\left(x,{\sqrt {ax^{2}+bx+c}}\right)\right)\,dx,}$$ 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 $${\displaystyle \int {\tilde {R}}_{1}(t)\log {\big (}{\tilde {R}}_{2}(t){\big )}\,dt,}$$ 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]

• Integration by substitution
• Trigonometric substitution
• Weierstrass substitution

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