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

Theorem

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

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:

$\ds \int \dfrac x {\sqrt {2 a x - x^2} } \rd x$

$=$

$\ds \int \dfrac {u + a} {\sqrt {a^2 - u^2} } \rd u$

$\ds $

$=$

$\ds \int \dfrac u {\sqrt {a^2 - u^2} } \rd u + a \int \dfrac 1 {\sqrt {a^2 - u^2} } \rd u$

$\ds $

$=$

$\ds -\sqrt {a^2 - u^2} + a \paren {\arcsin \frac x a}$

$\ds $

$=$

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

substituting for $u$ and simplifying

$\blacksquare$