Lemmata for Euler's Third Substitution/Lemma 4
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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:
- $t = \pm \sqrt {\dfrac {a \paren {x - \beta} } {x - \alpha} }$
Proof
\(\ds x\) | \(=\) | \(\ds \dfrac {a \beta - \alpha t^2} {a - t^2}\) | Lemmata for Euler's Third Substitution: Lemma $1$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \paren {a - t^2}\) | \(=\) | \(\ds a \beta - \alpha t^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds t^2 \paren {x - \alpha}\) | \(=\) | \(\ds a \paren {x - \beta}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds t^2\) | \(=\) | \(\ds \dfrac {a \paren {x - \beta} } {x - \alpha}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds t\) | \(=\) | \(\ds \pm \sqrt {\dfrac {a \paren {x - \beta} } {x - \alpha} }\) |
$\blacksquare$
Also see
- Results about Euler substitutions can be found here.