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

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Theorem

$\ds \int \paren {\sqrt {2 a x - x^2} }^n \rd x = \frac {\paren {x - a} \paren {\sqrt {2 a x - x^2} }^n} {n + 1} + \frac {n a^2} {n + 1} \int \paren {\sqrt {2 a x - x^2} }^{n - 2} \rd x$


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:

\(\ds \int \paren {\sqrt {2 a x - x^2} }^n \rd x\) \(=\) \(\ds \int \paren {\sqrt {a^2 - u^2} }^n \rd u\)
\(\ds \) \(=\) \(\ds \dfrac {u \paren {\sqrt {a^2 - u^2} }^n} {n + 1} - \dfrac {n a^2} {n + 1} \int \paren {\sqrt {a^2 - u^2} }^{n - 2} \rd u\) Primitive of $\paren {\sqrt {a^2 - u^2} }^n$
\(\ds \) \(=\) \(\ds \frac {\paren {x - a} \paren {\sqrt {2 a x - x^2} }^n} {n + 1} + \frac {n a^2} {n + 1} \int \paren {\sqrt {2 a x - x^2} }^{n - 2} \rd x\) substituting for $u$ and simplifying

$\blacksquare$


Sources

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