# Euler’s substitutions for integration

In the integration task

 $\int\!R(x,\,\sqrt{ax^{2}+bx+c})\,dx,$

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

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

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

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

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

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

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 $\sqrt{ax^{2}\!+\!bx\!+\!c}$ 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

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

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

 $\int\!\frac{dx}{\sqrt{x^{2}+1}}\;=\;\operatorname{arsinh}{x}+C,\quad\int\!% \frac{dx}{\sqrt{x^{2}-1}}\;=\;\operatorname{arcosh}{x}+C\;\;(x>1).$

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

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

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

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

And we obtain

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

## References

• 1 N. Piskunov: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Viies, täiendatud trükk.  Kirjastus “Valgus”, Tallinn (1965).
 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