Primitive of Root of x squared minus a squared cubed

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Theorem

$\ds \int \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$

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
\(\ds \leadsto \ \ \) \(\ds \int \paren {\sqrt {x^2 - a^2} }^3 \rd x\) \(=\) \(\ds \int \frac {\paren {\sqrt {z - a^2} }^3} {2 \sqrt z} \rd x\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {\sqrt z \paren {\sqrt {z - a^2} }^3} 4 - \frac {3 a^2} 8 \int \frac {\sqrt {z - a^2} } {\sqrt z} \rd x + C\) Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$
\(\ds \) \(=\) \(\ds \frac {\sqrt z \paren {\sqrt {z - a^2} }^3} 4 - \frac {3 a^2} 8 \paren {\sqrt z \sqrt {z - a^2} - \frac {a^2} 2 \int \frac {\d x} {\sqrt z \sqrt {z - a^2} } } + C\) Primitive of $\dfrac {\sqrt {p x + q} } {\sqrt{a x + b} }$
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2} 8 x \sqrt {x^2 + a^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {x^2 - a^2} } + C\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C\) Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$

$\blacksquare$


Also see


Sources

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