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[parent] Euler's substitutions for integration (Topic)

In the integration task

$\displaystyle \int\!R(x,\,\sqrt{ax^2+bx+c})\,dx,$
where the integrand is a rational function of $ x$ and $ \sqrt{ax^2+bx+c}$, the integrand can be changed to a rational function of a new variable $ t$ by using the following substitutions of Euler.
  • The first substitution of Euler. If $ a > 0$, we may write
    $\displaystyle \sqrt{ax^2+bx+c} = \pm x\sqrt{a}+t.$ (1)

    When we take $ \sqrt{a}$ with the minus sign, then
    $\displaystyle ax^2+bx+c = ax^2-2xt\sqrt{a}+t^2,$
    from which we get the expression
    $\displaystyle x = \frac{t^2-c}{b+2t\sqrt{a}};$
    thus also $ dx$ is expressible rationally via $ t$. We have
    $\displaystyle \sqrt{ax^2+bx+c} = -x\sqrt{a}+t = \frac{c-t^2}{b+2t\sqrt{a}}\sqrt{a}+t.$
  • The second substitution of Euler. If $ c > 0$, we take
    $\displaystyle \sqrt{ax^2+bx+c} = xt\pm\sqrt{c}.$ (2)

    With the minus sign we obtain, similarly as above,
    $\displaystyle x = \frac{2t\sqrt{c}+b}{t^2-a}.$
  • The third substitution of Euler. If the polynomial $ ax^2\!+\!bx\!+\!c$ has the real zeros $ \alpha$ and $ \beta$, we may chose
    $\displaystyle \sqrt{ax^2+bx+c} = (x-\alpha)t.$ (3)

    Now
    $\displaystyle ax^2+bx+c = a(x-\alpha)(x-\beta) = (x-\alpha)^2t^2,$
    whence $ a(x-\beta) = (x-\alpha)t^2$. This gives the expression
    $\displaystyle x = \frac{a\beta-\alpha t^2}{a-t^2}.$
    As in the preceding cases, we can express $ dx$ and $ \sqrt{ax^2\!+\!bx\!+\!c}$ rationally via $ t$.

Examples.

1. In the integral $ \displaystyle\int\!\frac{dx}{\sqrt{x^2+c}}$ we can use the first substitution: $ \sqrt{x^2+c} = -x+t$; then $ x^2+c = x^2-2xt+t^2$ and thus

$\displaystyle x = \frac{t^2-c}{2t},\;\; dx = \frac{t^2+c}{2t^2}\,dt,\;\; \sqrt{x^2+c} = -\frac{t^2-c}{2t}+t = \frac{t^2+c}{2t}.$
Accordingly we obtain
$\displaystyle \int\!\frac{dx}{\sqrt{x^2+c}} = \int\!\frac{\frac{t^2+c}{2t^2}dt}... ...2t}} = \int\!\frac{dt}{t} = \ln\vert t\vert+C = \ln\vert x+\sqrt{x^2+c}\vert+C.$
Especially the cases $ c = \pm1$ give the formulas
$\displaystyle \int\!\frac{dx}{\sqrt{x^2+1}} = {\mathrm{arsinh}}{x}+C, \quad \int\!\frac{dx}{\sqrt{x^2-1}} = {\mathrm{arcosh}}{x}+C \;\; (x > 1).$

2. The integral $ \displaystyle\int\!\frac{\sqrt{c^2-x^2}}{x}\,dx$ is needed in deriving the equation of the tractrix. We use for integrating the second substitution $ \sqrt{c^2-x^2} = xt-c$; then $ c^2-x^2 = x^2t^2-2cxt+c^2$, which implies

$\displaystyle x = \frac{2ct}{t^2+1},\;\; dx = \frac{2c(1-t^2)dt}{(1+t^2)^2},\;\; \sqrt{c^2-x^2} = \frac{2ct^2}{t^2+1}-c = \frac{c(t^2-1)}{t^2+1}.$
We then obtain
$\displaystyle \int\!\frac{\sqrt{c^2-x^2}}{x}\,dx = -c\int\!\frac{(1-t^2)^2}{t(1... ...c{4t}{(1+t^2)^2}-\frac{1}{t}\right)dt = -\frac{2c}{1+t^2}-c\ln\vert t\vert+C_1.$
The equation tying $ x$ and $ t$ gives $ \frac{2c}{1+t^2} = \frac{x}{t}$ and $ t = \frac{c+\sqrt{c^2-x^2}}{x}$, whence
$\displaystyle \int\!\frac{\sqrt{c^2-x^2}}{x}\,dx = -\frac{x^2}{c+\sqrt{c^2-x^2}... ...+\sqrt{c^2-x^2}}{x}+C_1 = -c+\sqrt{c^2-x^2}-c\ln\frac{c+\sqrt{c^2-x^2}}{x}+C_1,$
i.e.
$\displaystyle \int\!\frac{\sqrt{c^2-x^2}}{x}\,dx = \sqrt{c^2-x^2}-c\ln\frac{c+\sqrt{c^2-x^2}}{x}+C.$

3. In the integral $ \displaystyle\int\!\frac{dx}{\sqrt{x^2+3x-4}}$, the radicand is $ (x+4)(x-1)$. Using the third substitution of Euler, we take $ \sqrt{x^2+4x-3} = (x+4)t$. This simplifies to $ x-1 = (x+4)t^2$. Then we get

$\displaystyle x = \frac{1+4t^2}{1-t^2},\;\; dx = \frac{10t}{(1-t^2)^2}\,dt,\;\; \sqrt{x^2+3x-4} = \left(\frac{1+4t^^2}{1-t^2}+4\right)t = \frac{5t}{1-t^2}.$
And we obtain
$\displaystyle \int\!\frac{dx}{\sqrt{x^2+3x-4}} = \int\!\frac{10t(1-t^2)}{(1-t^2... ...left\vert\frac{1+\sqrt{\frac{x-1}{x+4}}}{1-\sqrt{\frac{x-1}{x+4}}}\right\vert+C$
$\displaystyle = \ln\left\vert\frac{\sqrt{x+4}+\sqrt{x-1}}{\sqrt{x+4}-\sqrt{x-1}}\right\vert+C.$

Bibliography

1
N. PISKUNOV: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Viies, täiendatud trükk. Kirjastus ``Valgus'', Tallinn (1965).



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See Also: integration of rational function of sine and cosine, tractrix

Other names:  integration of expressions of square roots of quadratic polynomials
Also defines:  Euler's substitutions, substitutions of Euler
Keywords:  square root of quadratic polynomial

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Cross-references: radicand, implies, tractrix, equation, integral, real, polynomial, expressible, expression, variable, rational function
There are 3 references to this entry.

This is version 11 of Euler's substitutions for integration, born on 2007-06-26, modified 2008-06-05.
Object id is 9681, canonical name is EulersSubstitutionsForIntegration.
Accessed 1743 times total.

Classification:
AMS MSC: 26A36 (Real functions :: Functions of one variable :: Antidifferentiation)
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