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

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Theorem

$\ds \int x^3 \sqrt {x^2 - a^2} \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C$

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
\(\ds \leadsto \ \ \) \(\ds \int x^3 \sqrt {x^2 - a^2} \rd x\) \(=\) \(\ds \int \frac {z^{3/2} \sqrt {z - a^2} \rd z} {2 \sqrt z}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int z \sqrt {z - a^2} \rd z\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 \paren {3 z + 2 a^2} } {15} \paren {\sqrt {z - a^2} }^3 } + C\) Primitive of $x \sqrt {\paren {a x + b} }$
\(\ds \) \(=\) \(\ds \frac {3 z + 2 a^2} {15} \paren {\sqrt {z - a^2} }^3 + C\) simplification
\(\ds \) \(=\) \(\ds \frac {3 x^2 + 2 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac {3 x^2 - 3 a^2 + 5 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C\)
\(\ds \) \(=\) \(\ds \frac {3 \paren {x^2 - a^2} \paren {\sqrt {a^2 - x^2} }^3 + 5 a^2 \paren {\sqrt {a^2 - x^2} }^3} {15} + C\)
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C\) simplification

$\blacksquare$


Also see


Sources