Primitive of Reciprocal of Root of a x + b by Root of p x + q/a greater than 0, p less than 0

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $a > 0$ and $p < 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C$

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


Proof

Completing the Square

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}

\ds \frac 1 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {b p - a q}^2} } & : a p > 0 \\ \ds \frac 1 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {b p - a q}^2 - u^2} } & : a p < 0 \end {cases}$

where:

$u := 2 a p x + b p + a q$

$\Box$


Hence:

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \frac 1 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {b p - a q}^2 - u^2} }\) Completing the Square above, as $a p < 0$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {-a p} } \map \arcsin {\dfrac u {b p - a q} } + C\) Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {b p - a q} } + C\) substituting for $u$ from Completing the Square above
We could finish here, but we wish to present the result in the format given, so:
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt {-a p} } \map \arcsin {-\dfrac {2 a p x + b p + a q} {b p - a q} } + C\) Arcsine is Odd Function
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C\) simplifying

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_Root_of_a_x_%2B_b_by_Root_of_p_x_%2B_q/a_greater_than_0,_p_less_than_0&oldid=630587"