Primitive of Reciprocal of Root of a minus x by Cube of Root of x minus b

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Theorem

$\ds \int \dfrac {\d x} {\paren {a - x}^{1/2} \paren {x - b}^{3/2} } = \dfrac 2 {b - a} \sqrt {\dfrac {a - x} {x - b} } + C$


Proof 1

Let:

\(\ds x\) \(=\) \(\ds a \cos^2 \theta + b \sin^2 \theta\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d x} {\d \theta}\) \(=\) \(\ds 2 a \cos \theta \paren {-\sin \theta} + 2 b \sin \theta \cos \theta\) Chain Rule for Derivatives, Derivative of Cosine Function, Derivative of Sine Function
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds 2 \paren {b - a} \cos \theta \sin \theta\)


Then we have:


\(\ds x\) \(=\) \(\ds a \cos^2 \theta + b \sin^2 \theta\)
\(\ds \) \(=\) \(\ds a \paren {1 - \sin^2 \theta} + b \sin^2 \theta\) Sum of Squares of Sine and Cosine
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds a - x\) \(=\) \(\ds \paren {a - b} \sin^2 \theta\) after algebra


and:

\(\ds x\) \(=\) \(\ds a \cos^2 \theta + b \sin^2 \theta\)
\(\ds \) \(=\) \(\ds a \cos^2 \theta + b \paren {1 - \cos^2 \theta}\) Sum of Squares of Sine and Cosine
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds x - b\) \(=\) \(\ds \paren {a - b} \cos^2 \theta\) after algebra
\(\text {(4)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {a - x} {x - b}\) \(=\) \(\ds \dfrac {\paren {a - b} \sin^2 \theta} {\paren {a - b} \cos^2 \theta}\) $(2)$ divided by $(3)$
\(\ds \dfrac {a - x} {x - b}\) \(=\) \(\ds \tan^2 \theta\) Definition of Real Tangent Function


Hence:

\(\ds \int \dfrac {\d x} {\paren {a - x}^{1/2} \paren {x - b}^{3/2} }\) \(=\) \(\ds \int \dfrac {2 \paren {b - a} \cos \theta \sin \theta \rd \theta} {\paren {a - b}^{1/2} \sin \theta \paren {a - b}^{3/2} \cos \theta}\) substituting from $(1)$, $(2)$ and $(3)$
\(\ds \) \(=\) \(\ds \dfrac 2 {b - a} \int \sec^2 \theta \rd \theta\) simplification, and Definition of Real Secant Function
\(\ds \) \(=\) \(\ds \dfrac 2 {b - a} \tan \theta + C\) Primitive of Square of Secant Function
\(\ds \) \(=\) \(\ds \dfrac 2 {b - a} \sqrt {\dfrac {a - x} {x - b} } + C\) substituting from $(4)$

$\blacksquare$


Proof 2

\(\ds \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C\) Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\ds \leadsto \ \ \) \(\ds \int \dfrac {\d x} {\paren {a - x}^{1/2} \paren {x - b}^{3/2} }\) \(=\) \(\ds \frac {2 \sqrt {\paren {-1} x + a} } {\paren {\paren {-1} \paren {-b} - a \cdot 1} \sqrt {1 \cdot x + \paren {-b} } } + C\) setting $p \gets 1$, $q \gets b$, $a \gets -1$, $b \gets a$
\(\ds \) \(=\) \(\ds \frac {2 \sqrt {a - x} } {\paren {b - a} \sqrt {x - b} } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 2 {b - a} \sqrt {\dfrac {a - x} {x - b} } + C\) rearranging

$\blacksquare$