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

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Theorem

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

for $\size x \ge a$.


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\paren {\sqrt {x^2 - a^2} }^3} x \rd x\) \(=\) \(\ds \int \frac {\paren {\sqrt {z - a^2} }^3} {2 z} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {z - a^2} }^3} 3 - \frac {a^2} 2 \int \frac {\sqrt {z - a^2} } z \rd z\) Primitive of $\dfrac {\paren {\sqrt {a x + b} }^m} x$
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {z - a^2} }^3} 3 - \frac {a^2} 2 \paren {2 \sqrt {z - a^2} - a^2 \int \frac {\d z} {z \sqrt {z - a^2} } }\) Primitive of $\dfrac {\sqrt {a x + b} } x$
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 - a^2 \sqrt {x^2 - a^2} + a^4 \int \frac {\d x} {x \sqrt {x^2 - a^2} }\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 - a^2 \sqrt {x^2 - a^2} + a^3 \arcsec \size {\frac x a} + C\) Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$

$\blacksquare$


Also see


Sources

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