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

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Theorem

$\ds \int x^2 \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {-x \paren {\sqrt {a^2 - x^2} }^5} 6 + \frac {a^2 x \paren {\sqrt {a^2 - x^2} }^3} {24} + \frac {a^4 x \sqrt {a^2 - x^2} } {16} + \frac {a^6} {16} \arcsin \frac x a + C$


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int x^2 \paren {\sqrt {a^2 - x^2} }^3 \rd x\) \(=\) \(\ds \int \frac {\sqrt z \paren {a^2 - z}^{3/2} } 2 \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {\sqrt z \paren {a^2 - z}^{5/2} } {-6} + \frac {-a^2} {-12} \int \frac {\paren {a^2 - z}^{3/2} } {\sqrt z} \rd z + C\) Primitive of $\paren {p x + q}^n \sqrt{a x + b}$
\(\ds \) \(=\) \(\ds \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {a^2 - x^2}^{3/2} \rd x + C\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {-x \paren {a^2 - x^2}^{5/2} } 6\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a} + C\) Primitive of $\paren {\sqrt {a^2 - x^2} }^3$
\(\ds \) \(=\) \(\ds \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {a^2 - x^2} }^3} {24}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {a^4 x \sqrt {a^2 - x^2} } {16} + \frac {a^6} {16} \arcsin \frac x a + C\) simplifying

$\blacksquare$


Also see


Sources

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