Primitive of Reciprocal of Root of 2 a x minus x squared

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Theorem

$\ds \int \frac {\d x} {\sqrt {2 a x - x^2} } = \arcsin \dfrac {x - a} a + C$

where $C$ is an arbitrary constant.


Proof

Let $u := x - a$.

Then:

$\dfrac {\d u} {\d x} = 1$

and:

$x = u + a$

and we have:

\(\ds \int \frac {\d x} {\sqrt {2 a x - x^2} }\) \(=\) \(\ds \int \frac {\d u} {\sqrt {2 a \paren {u + a} - \paren {u + a}^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac {\d u} {\sqrt {2 a u + 2 a^2 - u^2 - 2 a u - a^2} }\) multiplying out
\(\ds \) \(=\) \(\ds \int \frac {\d u} {\sqrt {a^2 - u^2} }\) simplifying
\(\ds \) \(=\) \(\ds \arcsin \frac u a + C\) Primitive of $\dfrac 1 {\sqrt {a^2 - u^2} }$: Arcsine Form
\(\ds \) \(=\) \(\ds \arcsin \dfrac {x - a} a + C\) substituting for $u$

$\blacksquare$


Sources

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