Английская Википедия:Euler substitution
Шаблон:Short description Шаблон:Calculus Euler substitution is a method for evaluating integrals of the form
<math display="block">\int R(x, \sqrt{ax^2 + bx + c}) \, dx,</math>
where <math>R</math> is a rational function of <math>x</math> and <math display="inline">\sqrt{ax^2 + bx + c}</math>. 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 <math>a > 0</math>. We substitute <math display="block"> \sqrt{ax^2 + bx + c} = \pm x\sqrt{a} + t</math> and solve the resulting expression for <math>x</math>. We have that <math>x = \frac{c - t^2}{\pm 2t\sqrt{a} - b}</math> and that the <math> dx</math> term is expressible rationally in <math>t</math>.
In this substitution, either the positive sign or the negative sign can be chosen.
Euler's second substitution
If <math>c > 0</math>, we take <math display="block"> \sqrt{ax^2 + bx + c} = xt \pm \sqrt{c}.</math> We solve for <math>x</math> similarly as above and find <math display="block">x = \frac{\pm 2t\sqrt{c} - b}{a - t^2}.</math>
Again, either the positive or the negative sign can be chosen.
Euler's third substitution
If the polynomial <math>ax^2 + bx + c</math> has real roots <math>\alpha</math> and <math>\beta</math>, we may choose <math display="inline">\sqrt{ax^2 + bx + c} = \sqrt{a(x - \alpha)(x - \beta)} = (x - \alpha)t</math>. This yields <math>x = \frac{a\beta - \alpha t^2}{a - t^2},</math> and as in the preceding cases, we can express the entire integrand rationally in <math>t</math>.
Worked examples
Examples for Euler's first substitution
One
In the integral <math>\int\! \frac{\ dx}{\sqrt{x^2+c}}</math> we can use the first substitution and set <math display="inline">\sqrt{x^2+c} = -x+t</math>, thus <math display="block">x = \frac{t^2-c}{2t} \quad\quad \ dx = \frac{t^2+c}{2t^2}\,\ dt</math> <math display="block">\sqrt{x^2+c} = -\frac{t^2-c}{2t}+t = \frac{t^2+c}{2t}</math> Accordingly, we obtain: <math display="block">\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</math>
The cases <math>c = \pm 1</math> give the formulas <math display="block"> \begin{align} \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{align} </math>
Two
For finding the value of <math display="block">\int\frac{1}{x\sqrt{x^{2}+4x-4}}dx,</math> we find <math>t</math> using the first substitution of Euler, <math display="inline">\sqrt{x^{2}+4x-4} = \sqrt{1}x+t = x+t</math>. Squaring both sides of the equation gives us <math>x^{2}+4x-4 = x^{2} + 2xt +t^{2}</math>, from which the <math>x^2</math> terms will cancel out. Solving for <math>x</math> yields <math display="block">x=\frac{t^{2}+4}{4-2t}.</math>
From there, we find that the differentials <math>dx</math> and <math>dt</math> are related by <math>dx=\frac{-2t^{2}+8t+8}{(4-2t)^{2}}dt.</math>
Hence, <math display="block"> \begin{align} \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 t2\right) +C\\[6pt] &= \tan^{-1}\left(\frac{\sqrt{x^{2}+4x-4}-x}{2}\right)+C \end{align} </math>
Examples for Euler's second substitution
In the integral <math display="block">\int\! \frac{dx}{x\sqrt{-x^2+x+2}},</math> we can use the second substitution and set <math>\sqrt{-x^2+x+2} = xt + \sqrt{2}</math>. Thus <math display="block">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,</math> and <math display="block">\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}</math>
Accordingly, we obtain: <math display="block"> \begin{align} \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{align} </math>
Examples for Euler's third substitution
To evaluate <math display="block">\int\! \frac{x^2}{\sqrt{-x^2+3x-2}}\ dx,</math> we can use the third substitution and set <math display="inline">\sqrt{-(x-2)(x-1)} = (x-2)t</math>. Thus <math display="block">x = \frac{-2t^2-1}{-t^2-1} \qquad \ dx = \frac{2t}{(-t^2-1)^2}\,\ dt,</math> and <math display="block">\sqrt{-x^2+3x-2} = (x-2)t = \frac{t}{-t^2-1.}</math>
Next, <math display="block">\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.</math> 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 <math display="inline">\int \frac{dx}{\sqrt{-x^2 + c}}</math>, the substitution <math display="inline">\sqrt{-x^2 + c} = \pm ix + t</math> 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 <math display="block">\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,</math> where <math>R_1</math> and <math>R_2</math> are rational functions of <math>x</math> and <math display="inline">\sqrt{ax^2 + bx + c}</math>. This integral can be transformed by the substitution <math display="inline">\sqrt{ax^2 + bx + c} = \sqrt{a} + xt</math> into another integral <math display="block">\int \tilde R_1(t) \log\big(\tilde R_2(t)\big) \, dt,</math> where <math>\tilde R_1(t)</math> and <math>\tilde R_2(t)</math> are now simply rational functions of <math>t</math>. 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
References
Шаблон:Reflist Шаблон:PlanetMath attribution
- ↑ 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.
- ↑ Шаблон:Cite book
