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

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Theorem

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


Proof

Let $u := x - a$.

Then:

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

and:

$x = u + a$


Then:

\(\ds 2 a x - x^2\) \(=\) \(\ds 2 a \paren {u + a} - \paren {u + a}^2\)
\(\ds \) \(=\) \(\ds 2 a u + 2 a^2 - u^2 - 2 a u - a^2\)
\(\ds \) \(=\) \(\ds a^2 - u^2\)


and we have:

\(\ds \int \dfrac {\sqrt {2 a x - x^2} } x \rd x\) \(=\) \(\ds \int \dfrac {\sqrt {a^2 - u^2} } {u + a} \rd u\)
\(\ds \) \(=\) \(\ds \int \sqrt {\dfrac {a^2 - u^2} {\paren {a + u}^2} } \rd u\)
\(\ds \) \(=\) \(\ds \int \sqrt {\dfrac {a - u} {a + u} } \rd u\) Difference of Two Squares and cancelling $a + x$
\(\ds \) \(=\) \(\ds \sqrt {a^2 - u^2} + a \arcsin \dfrac u a + C\) result to be found
\(\ds \) \(=\) \(\ds \sqrt {2 a x - x^2} + a \arcsin \dfrac {x - a} a + C\) substituting for $u$ and simplifying


This needs considerable tedious hard slog to complete it.
In particular: There's a result in there somewhere.
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Sources

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