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

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Theorem

$\ds \int x \sqrt {2 a x - x^2} \rd x = \frac {\paren {x + a} \paren {2 x - 3 a} \sqrt {2 a x - x^2} } 6 + \frac {a^3} 2 \arcsin \dfrac {x - a} a + C$


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:

\(\ds \int x \sqrt {2 a x - x^2} \rd x\) \(=\) \(\ds \int \paren {u + a} \sqrt {a^2 - u^2} \rd u\)
\(\ds \) \(=\) \(\ds \int u \sqrt {a^2 - u^2} \rd u + a \int \sqrt {a^2 - u^2} \rd u\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {a^2 - u^2} }^3} 3 - a \int \sqrt {a^2 - u^2} \rd u\) Primitive of $u \sqrt {a^2 - u^2}$
\(\ds \) \(=\) \(\ds \frac {-\paren {a^2 - u^2} \sqrt {a^2 - u^2} } 3 - a \paren {\frac {u \sqrt {a^2 - u^2} } 2 + \frac {a^2} 2 \arcsin \frac u a} + C\) Primitive of $\sqrt {a^2 - u^2}$: Arcsine Form
\(\ds \) \(=\) \(\ds \frac {-\paren {2 a x - x^2} \sqrt {2 a x - x^2} } 3 - \frac {a \paren {x - a} \sqrt {2 a x - x^2} } 2 + \frac {a^3} 2 \arcsin \frac {x - a} a + C\) substituting for $u$ and simplifying

The result follows after algebra.


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$\blacksquare$


Sources

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