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 |
$\blacksquare$