Euler’s substitutions for integration

In the integration task

$$\int R(x,\sqrt{a{x}^2+bx+c})𝑑x,$$

where the integrand is a rational function of $x$ and $\sqrt{a{x}^2+bx+c}$, 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

  $\sqrt{a{x}^2+bx+c}=\pm x\sqrt{a}+t.$ (1)

  When we take $\sqrt{a}$ with the minus sign, then

  $$a{x}^2+bx+c=a{x}^2-2xt\sqrt{a}+t^2,$$

  from which we get the expression

  $$x=\frac{t^2-c}{b+2t\sqrt{a}};$$

  thus also $dx$ is expressible rationally via $t$. We have

  $$\sqrt{a{x}^2+bx+c}=-x\sqrt{a}+t=\frac{c-t^2}{b+2t\sqrt{a}}\sqrt{a}+t.$$

• The second substitution of Euler.  If  $c>0$,  we take

  $\sqrt{a{x}^2+bx+c}=xt\pm \sqrt{c}.$ (2)

  With the minus sign we obtain, similarly as above,

  $$x=\frac{2t\sqrt{c}+b}{t^2-a}.$$

• The third substitution of Euler.  If the polynomial $a{x}^2+bx+c$ has the real zeros $\alpha$ and $\beta$, we may chose

  $\sqrt{a{x}^2+bx+c}=(x-\alpha )t.$ (3)

  Now

  $$a{x}^2+bx+c=a(x-\alpha )(x-\beta )={(x-\alpha )}^2{t}^2,$$

  whence  $a(x-\beta )=(x-\alpha )t^2$. This gives the expression

  $$x=\frac{a\beta -\alpha t^2}{a-t^2}.$$

  As in the preceding cases, we can express $dx$ and $\sqrt{a{x}^2+bx+c}$ 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

$$x=\frac{t^2-c}{2t},dx=\frac{t^2+c}{2t^2}dt,\sqrt{x^2+c}=-\frac{t^2-c}{2t}+t=\frac{t^2+c}{2t}.$$

Accordingly we obtain

$$\int \frac{dx}{\sqrt{x^2+c}}=\int \frac{\frac{t^2+c}{2t^2}dt}{\frac{t^2+c}{2t}}=\int \frac{dt}{t}=\ln |t|+C=\ln |x+\sqrt{x^2+c}|+C.$$

Especially the cases  $c=\pm 1$  give the formulas

$$\int \frac{dx}{\sqrt{x^2+1}}=\mathrm{arsinh}x+C,\int \frac{dx}{\sqrt{x^2-1}}=\mathrm{arcosh}x+C(x>1).$$

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  $c^2-x^2=x^2{t}^2-2cxt+c^2$, which implies

$$x=\frac{2ct}{t^2+1},dx=\frac{2c(1-t^2)dt}{{(1+t^2)}^2},\sqrt{c^2-x^2}=\frac{2c{t}^2}{t^2+1}-c=\frac{c(t^2-1)}{t^2+1}.$$

We then obtain

$$\int \frac{\sqrt{c^2-x^2}}{x}𝑑x=-c\int \frac{{(1-t^2)}^2}{t{(1+t^2)}^2}𝑑t=c\int \left(\frac{4t}{{(1+t^2)}^2}-\frac{1}{t}\right)𝑑t=-\frac{2c}{1+t^2}-c\ln |t|+C_1.$$

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

$$\int \frac{\sqrt{c^2-x^2}}{x}𝑑x=-\frac{x^2}{c+\sqrt{c^2-x^2}}-c\ln \frac{c+\sqrt{c^2-x^2}}{x}+C_1=-c+\sqrt{c^2-x^2}-c\ln \frac{c+\sqrt{c^2-x^2}}{x}+C_1,$$

i.e.

$$\int \frac{\sqrt{c^2-x^2}}{x}𝑑x=\sqrt{c^2-x^2}-c\ln \frac{c+\sqrt{c^2-x^2}}{x}+C.$$

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

$$x=\frac{1+4t^2}{1-t^2},dx=\frac{10t}{{(1-t^2)}^2}dt,\sqrt{x^2+3x-4}=\left(\frac{1+4tr}{1-t^2}+4\right)t=\frac{5t}{1-t^2}.$$

And we obtain

$$\int \frac{dx}{\sqrt{x^2+3x-4}}=\int \frac{10t(1-t^2)}{{(1-t^2)}^2\cdot 5t}𝑑t=\int \frac{2}{1-t^2}𝑑t=\ln \left|\frac{1+t}{1-t}\right|+C=\ln \left|\frac{1+\sqrt{\frac{x-1}{x+4}}}{1-\sqrt{\frac{x-1}{x+4}}}\right|+C$$

$$=\ln \left|\frac{\sqrt{x+4}+\sqrt{x-1}}{\sqrt{x+4}-\sqrt{x-1}}\right|+C.$$

Title	Euler’s substitutions for integration
Canonical name	EulersSubstitutionsForIntegration
Date of creation	2013-03-22 17:19:43
Last modified on	2013-03-22 17:19:43
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	15
Author	pahio (2872)
Entry type	Topic
Classification	msc 26A36
Synonym	integration of expressions of square roots of quadratic polynomials
Related topic	IntegrationOfRationalFunctionOfSineAndCosine
Related topic	Tractrix
Related topic	Arsinh
Related topic	Arcosh
Defines	Euler’s substitutions
Defines	substitutions of Euler