Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0/Proof 2

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Theorem

Let $a, b, p, q \in \R$ such that $a p \ne b q$.

Let $a p < 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C$

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.


Proof

First let us express the integrand in the following form:

\(\text {(1)}: \quad\) \(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \int \frac {\d x} {\sqrt {a p \paren {x - \paren {-\frac b a} } \paren {x - \paren {-\frac q p} } } }\)


Recall the definition of Euler's third substitution:

Let $a x^2 + b x + c$ have real roots $\alpha$ and $\beta$.


Euler's third substitution is the substitution:

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

Then:

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

and hence $\d x$ is expressible as a rational function of $x$.


In this context we have:

\(\ds a\) \(\gets\) \(\ds a p\)
\(\ds \alpha\) \(\gets\) \(\ds -\frac b a\)
\(\ds \beta\) \(\gets\) \(\ds -\frac q p\)


Hence we make the substitution:

\(\ds x\) \(=\) \(\ds \dfrac {a p \paren {-\frac q p} - \paren {-\frac b a} t^2} {a p - t^2}\) Lemmata for Euler's Third Substitution: Lemma $1$
\(\ds \) \(=\) \(\ds \dfrac {-a q + \paren {\frac b a} t^2} {a p - t^2}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {a^2 q - b t^2} {a \paren {t^2 - a p} }\) multiplying top and bottom by $-a$


Then we use:

\(\ds \leadsto \ \ \) \(\ds x - \alpha\) \(=\) \(\ds \dfrac {a \paren {\alpha - \beta} } {t^2 - a}\) Lemmata for Euler's Third Substitution: Lemma $2$
\(\ds \) \(=\) \(\ds \dfrac {a p \paren {\paren {-\frac b a} - \paren {-\frac q p} } } {t^2 - a p}\) substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds x - \alpha\) \(=\) \(\ds \dfrac {a q - p b} {t^2 - a p}\) simplifying


Then:

\(\ds \dfrac {\d x} {\d t}\) \(=\) \(\ds \dfrac {2 t a \paren {\beta - \alpha} } {\paren {a - t^2}^2}\) Lemmata for Euler's Third Substitution: Lemma $3$
\(\ds \) \(=\) \(\ds \dfrac {2 t a p \paren {\paren {-\frac q p} - \paren {-\frac b a} } } {\paren {a p - t^2}^2}\) substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d x} {\d t}\) \(=\) \(\ds \dfrac {2 t \paren {b p - a q} } {\paren {t^2 - a p}^2}\) simplifying


Then:

\(\ds t\) \(=\) \(\ds \sqrt {\dfrac {a \paren {x - \beta} } {x - \alpha} }\) Lemmata for Euler's Third Substitution: Lemma $4$
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a p \paren {x - \paren {-\frac q p} } } {x - \paren {-\frac b a} } }\) substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a p \paren {x + \frac q p} } {x + \frac b a} }\) rearranging
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a^2 \paren {p x + q} } {a x + b} }\) rearranging
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds a \sqrt {\dfrac {p x + q} {a x + b} }\) simplifying


Assembling the pieces:

\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \int \frac {\d t} {\paren {\dfrac {a q - p b} {t^2 - a p} } t} \dfrac {2 t \paren {b p - a q} } {\paren {t^2 - a p}^2}\) Integration by Substitution, using $\sqrt {\paren {a x + b} \paren {p x + q} } = \paren {x - \alpha} t$
\(\ds \) \(=\) \(\ds \int \frac {t^2 - a p} {\paren {a q - p b} t} \cdot \dfrac {-2 t \paren {a q - b p} } {\paren {t^2 - a p}^2} \rd t\) simplifying
\(\ds \) \(=\) \(\ds -2 \int \dfrac {\d t} {\paren {t^2 - a p} }\) simplifying
\(\ds \) \(=\) \(\ds -2 \int \dfrac {\d t} {\paren {t^2 + \paren {-a p} } }\) as $a p < 0$ by hypothesis
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arctan {\dfrac t {\sqrt {-a p} } } + C\) Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arctan {\dfrac {a \sqrt {\frac {p x + q} {a x + b} } } {\sqrt {-a p} } } + C\) substituting for $t$ from $(2)$
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-a \paren {p x + q} } {p \paren {a x + b} } } } + C\)
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arccot {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C\) Arccotangent of Reciprocal equals Arctangent
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \paren {\dfrac \pi 2 - \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } } + C\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \paren {\map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } - \dfrac \pi 2} + C\) Arctangent is Odd Function
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C\) subsuming $\dfrac 2 {\sqrt {-a p} } \times \dfrac {-\pi} 2$ into the arbitrary constant

$\blacksquare$


Sources

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