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

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Theorem

$\ds \int x^3 \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {\paren {\sqrt {a^2 - x^2} }^7} 7 - \frac {a^2 \paren {\sqrt {a^2 - x^2} }^5} 5 + C$


Proof

\(\ds \int x^3 \paren {\sqrt {a^2 - x^2} }^3 \rd x\) \(=\) \(\ds \int x \paren {x^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x\)
\(\ds \) \(=\) \(\ds \int x \paren {x^2 - a^2 + a^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x\) Primitive of Power
\(\ds \) \(=\) \(\ds -\int x \paren {a^2 - x^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds -\int x \paren {\sqrt {a^2 - x^2} }^5 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x\) simplifying


Let:

\(\ds z\) \(=\) \(\ds a^2 - x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds -2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds -\int x \paren {\sqrt {a^2 - x^2} }^5 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x\)
\(\ds \) \(=\) \(\ds -\int \frac {\sqrt z z^{5/2} } {-2 \sqrt z} \rd z + a^2 \int \frac {\sqrt z z^{3/2} } {-2 \sqrt z} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int z^{5/2} \rd z - \frac {a^2} 2 \int z^{3/2} \rd z\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 2 \frac {z^{7/2} } {7/2} - \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {z^{7/2} } 7 - a^2 \frac {z^{5/2} } 5 + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {a^2 - x^2} }^7} 7 - \frac {a^2 \paren {\sqrt {a^2 - x^2} }^5} 5 + C\) substituting for $z$

$\blacksquare$


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Sources