|
|
| Line 1: |
Line 1: |
| '''Euler substitution''' is a method for evaluating integrals of the form:
| | Electorate Officer Emerson from Geraldton, has several hobbies and interests including hang gliding, airsoft and aircraft spotting. Last month just arrived at Iguaçu National Park.<br><br>Feel free to surf to my blog post :: [http://ganhandodinheironainternet.comoganhardinheiro101.com/ como ganhar dinheiro na internet] |
| | |
| :<math>\int\!R(x,\sqrt{ax^2+bx+c})dx </math>
| |
| | |
| where <math>R</math> is a rational function of <math>x</math> and <math>\sqrt{ax^2+bx+c}</math>. In such cases the integrand can be changed to a rational function of a new variable <math>t</math> by using the following substitutions of Euler:<ref>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.</ref>
| |
| | |
| ==The first substitution of Euler a > 0==
| |
| If a > 0 we may write
| |
| <math>
| |
| \sqrt{ax^2+bx+c} \;=\; \pm x\sqrt{a}+t. | |
| </math> | |
| When we take <math>\sqrt{a}</math> with the minus sign, then
| |
| <math>ax^2+bx+c \;=\; ax^2-2xt\sqrt{a}+t^2</math>
| |
| from which we get the expression
| |
| <math>x = \frac{t^2-c}{b+2t\sqrt{a}}</math>
| |
| thus also <math>dx</math> is expressible rationally via <math>t</math> . We have
| |
| <math>\sqrt{ax^2+bx+c} \;=\; -x\sqrt{a}+t \;=\; \frac{c-t^2}{b+2t\sqrt{a}}\sqrt{a}+t</math>.
| |
| | |
| ==The second substitution of Euler. c > 0==
| |
| If c > 0 we take
| |
| <math>
| |
| \begin{align}
| |
| \sqrt{ax^2+bx+c} \;=\; xt\pm\sqrt{c}.
| |
| \end{align}
| |
| </math>
| |
| With the minus sign we obtain, similarly as above,
| |
| <math>x \;=\; \frac{2t\sqrt{c}+b}{t^2-a}.</math>
| |
| | |
| ==The third substitution of Euler==
| |
| If the polynomial <math>ax^2 + bx + c</math> has the real zeros <math>\alpha</math> and <math>\beta</math>, we may chose
| |
| :<math>\sqrt{ax^2\!+\!bx\!+\!c} \;=\; (x\!-\!\alpha)t</math>
| |
| Now
| |
| :<math>ax^2\!+\!bx\!+\!c \;=\; a(x\!-\!\alpha)(x\!-\!\beta) \;=\; (x\!-\!\alpha)^2t^2</math> | |
| :<math>a(x\!-\!\beta) = (x\!-\!\alpha)t^2</math> | |
| This gives the expression
| |
| :<math>x = \frac{a\beta-\alpha t^2}{a-t^2}</math> | |
| As in the preceding cases, we can express <math>dx</math> and <math>\sqrt{ax^2\!+\!bx\!+\!c}</math> rationally via <math>t</math>.
| |
| | |
| ==Examples==
| |
| {{Expand section|date=March 2012}}
| |
| In the integral <math>\int\ \frac{dx}{\sqrt{x^2+c}}</math> we can use the first substitution:<math>\sqrt{x^2+c} = -x+t</math>, then <math>x^2+c = x^2-2xt+t^2</math> and thus
| |
| :<math>x = \frac{t^2-c}{2t} \quad\quad dx = \frac{t^2+c}{2t^2} dt</math>
| |
| :<math>\sqrt{x^2+c} = -\frac{t^2-c}{2t}+t = \frac{t^2+c}{2t}</math>
| |
| Accordingly we obtain:
| |
| :<math>\int \frac{dx}{\sqrt{x^2+c}} = \int \frac{\frac{t^2+c}{2t^2}dt}{\frac{t^2+c}{2t}}</math>
| |
| :<math> = \int\!\frac{dt}{t} = \ln|t|+C = \ln|x+\sqrt{x^2+c}|+C</math>
| |
| Especially the cases <math>c = \pm 1</math>, give the formulas
| |
| :<math>\int \frac{dx}{\sqrt{x^2+1}} = \mbox{arcsinh}(x) + C </math>
| |
| :<math>\int \frac{dx}{\sqrt{x^2-1}} = \mbox{arccosh}(x) + C \quad (x > 1)</math>
| |
| | |
| ==References==
| |
| {{reflist}}
| |
| {{PlanetMath attribution|id=9681|title=Eulers Substitutions For Integration}}
| |
| | |
| [[Category:Integral calculus]]
| |
Electorate Officer Emerson from Geraldton, has several hobbies and interests including hang gliding, airsoft and aircraft spotting. Last month just arrived at Iguaçu National Park.
Feel free to surf to my blog post :: como ganhar dinheiro na internet