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

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Theorem

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


Proof

With a view to expressing the problem in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 1\) Power Rule for Derivatives


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac x {\paren {\sqrt {a^2 - x^2} }^3}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac 1 {\sqrt {a^2 - x^2} }\) Primitive of $\dfrac x {\paren {\sqrt {a^2 - x^2} }^3}$


Then:

\(\ds \int \frac {x^2 \rd x} {\paren {\sqrt {a^2 - x^2} }^3}\) \(=\) \(\ds \int x \frac {x \rd x} {\paren {\sqrt {a^2 - x^2} }^3}\)
\(\ds \) \(=\) \(\ds x \frac 1 {\sqrt {a^2 - x^2} } - \int \frac 1 {\sqrt {a^2 - x^2} } 1 + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac x {\sqrt {a^2 - x^2} } - \int \frac 1 {\sqrt {a^2 - x^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac x {\sqrt {a^2 - x^2} } - \arcsin \frac x a + C\) Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$

$\blacksquare$


Also see


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