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

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Theorem

\(\ds \int \frac {x^2 \rd x} {\sqrt {a^2 - x^2} }\) \(=\) \(\ds \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C\)
\(\ds \) \(=\) \(\ds \frac {a^2} 2 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} } 2 + C\) either way round, whichever you prefer


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:

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


Then:

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

$\blacksquare$


Also see


Sources