Euler substitution

Euler's second substitution

If ${\displaystyle c>0}$, we take $${\displaystyle \sqrt{a{x}^2+bx+c}=xt\pm \sqrt{c}.}$$ 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.

Euler's third substitution

If the polynomial ${\displaystyle a{x}^2+bx+c}$ has real roots ${\displaystyle \alpha }$ and ${\displaystyle \beta }$, we may choose $\sqrt{a{x}^2+bx+c}=\sqrt{a(x-\alpha )(x-\beta )}=(x-\alpha )t$. 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 ${\displaystyle t}$.

Examples

Examples for Euler's first substitution

First example

For the integral ${\displaystyle \int \frac{dx}{\sqrt{x^2+c}}}$, we can use the first substitution and set $\sqrt{x^2+c}=-x+t$. Thus, $${\displaystyle x=\frac{t^2-c}{2t},dx=\frac{t^2+c}{2t^2}dt,}$$ $${\displaystyle \sqrt{x^2+c}=-\frac{t^2-c}{2t}+t=\frac{t^2+c}{2t}.}$$ 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 $${\displaystyle \begin{array}{cc}\int \frac{dx}{\sqrt{x^2+1}}& =\mathrm{arsinh}(x)+C,\\ \int \frac{dx}{\sqrt{x^2-1}}& =\mathrm{arcosh}(x)+C.(x>1)\end{array}}$$

Second example

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: $\sqrt{x^2+4x-4}=\sqrt{1}x+t=x+t$. 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, $${\displaystyle \begin{array}{cccc}\int \frac{dx}{x\sqrt{x^2+4x-4}}& =\int \frac{\frac{-2t^2+8t+8}{(4-2t{)}^2}}{\left(\frac{t^2+4}{4-2t}\right)\left(\frac{-t^2+4t+4}{4-2t}\right)}dt& & t=\sqrt{x^2+4x-4}-x\\ & =2\int \frac{dt}{t^2+4}={\tan }^{-1}\left(\frac{t}{2}\right)+C\\ & ={\tan }^{-1}\left(\frac{\sqrt{x^2+4x-4}-x}{2}\right)+C.\end{array}}$$

Examples for Euler's second substitution

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 $${\displaystyle x=\frac{1-2\sqrt{2}t}{t^2+1},dx=\frac{2\sqrt{2}{t}^2-2t-2\sqrt{2}}{(t^2+1{)}^2}dt,}$$ 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: $${\displaystyle \begin{array}{cc}\int \frac{dx}{x\sqrt{-x^2+x+2}}& =\int \frac{\frac{2\sqrt{2}{t}^2-2t-2\sqrt{2}}{(t^2+1{)}^2}}{\frac{1-2\sqrt{2}t}{t^2+1}\frac{-\sqrt{2}{t}^2+t+\sqrt{2}}{t^2+1}}dt\\ & =\int \frac{-2}{-2\sqrt{2}t+1}dt=\frac{1}{\sqrt{2}}\int \frac{-2\sqrt{2}}{-2\sqrt{2}t+1}dt\\ & =\frac{1}{\sqrt{2}}\ln \left|2\sqrt{2}t-1\right|+C\\ & =\frac{\sqrt{2}}{2}\ln \left|2\sqrt{2}\frac{\sqrt{-x^2+x+2}-\sqrt{2}}{x}-1\right|+C.\end{array}}$$

Examples for Euler's third substitution

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

Next, $${\displaystyle \int \frac{x^2}{\sqrt{-x^2+3x-2}}dx=\int \frac{{\left(\frac{-2t^2-1}{-t^2-1}\right)}^2\frac{2t}{(-t^2-1{)}^2}}{\frac{t}{-t^2-1}}dt=\int \frac{2(-2t^2-1{)}^2}{(-t^2-1{)}^3}dt.}$$ 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 $\int \frac{dx}{\sqrt{-x^2+c}}$, 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 $${\displaystyle \int R_1\left(x,\sqrt{a{x}^2+bx+c}\right)\log \left(R_2\left(x,\sqrt{a{x}^2+bx+c}\right)\right)dx,}$$ where ${\displaystyle R_1}$ and ${\displaystyle R_2}$ are rational functions of ${\displaystyle x}$ and $\sqrt{a{x}^2+bx+c}$. This integral can be transformed by the substitution $\sqrt{a{x}^2+bx+c}=\sqrt{a}+xt$ into another integral $${\displaystyle \int {\tilde{R}}_1(t)\log ({\tilde{R}}_2(t))dt,}$$ 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.