Primitive of Reciprocal of x by Root of 2 a x minus x squared
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Theorem
- $\ds \int \dfrac 1 {x \sqrt {2 a x - x^2} } \rd x = -\dfrac 1 a \sqrt {\dfrac {2 a - x} x} + C$
Proof
Recall Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$:
- $\ds \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C$
Then:
| \(\ds \int \dfrac 1 {x \sqrt {2 a x - x^2} } \rd x\) | \(=\) | \(\ds \int \dfrac 1 {x \sqrt {x \paren {2 a - x} } } \rd x\) | manipulating subject into correct form | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \sqrt {2 a - x} } {\paren {\paren {-1} \times 0 - 2 a \times 1} \sqrt {1 \times x + 0} } + C\) | $p \gets 1$, $q \gets 0$, $a \gets -1$, $b \gets 2 a$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \sqrt {2 a - x} } {\paren {-2 a} \sqrt x} + C\) | simplification | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac 1 a \sqrt {\dfrac {2 a - x} x} + C\) | simplification |
$\blacksquare$
Sources
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $55$.