Lemmata for Euler's Third Substitution/Lemma 1

Theorem

Let Euler's third substitution be employed in order to evaluate a primitive of the form:

$\ds \map R {x, \sqrt {a x^2 + b x + c} } \rd x$

where $R$ is a rational function of $x$ and $\sqrt {a x^2 + b x + c}$.

Thus:

$\ds \sqrt {a x^2 + b x + c} = \sqrt {a \paren {x - \alpha} \paren {x - \beta} } = \paren {x - \alpha} t$

Then we have:

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

Proof

\(\ds \sqrt {a x^2 + b x + c}\)

\(=\)

\(\ds \sqrt {a \paren {x - \alpha} \paren {x - \beta} }\)

\(\ds \)

\(=\)

\(\ds \paren {x - \alpha} t\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds a \paren {x - \alpha} \paren {x - \beta}\)

\(=\)

\(\ds \paren {x - \alpha}^2 t^2\)

squaring both sides

\(\ds \leadsto \ \ \)

\(\ds a \paren {x - \beta}\)

\(=\)

\(\ds \paren {x - \alpha} t^2\)

simplifying

\(\ds \leadsto \ \ \)

\(\ds a x - a \beta\)

\(=\)

\(\ds x t^2 - \alpha t^2\)

multiplying out

\(\ds \leadsto \ \ \)

\(\ds x \paren {a - t^2}\)

\(=\)

\(\ds a \beta - \alpha t^2\)

rearranging

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \dfrac {a \beta - \alpha t^2} {a - t^2}\)

rearranging

$\blacksquare$

Also see

• Definition:Euler's Third Substitution