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

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Theorem

$\displaystyle \int x^2 \sqrt {a^2 - x^2} \rd x = \frac {-x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \sinh^{-1} \frac x a + C$


Proof

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int x^2 \sqrt {a^2 - x^2} \rd x\) \(=\) \(\ds \int \frac {z \sqrt {a^2 - z} \rd z} {2 \sqrt z}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \sqrt z \sqrt {a^2 - z} \rd z\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {-2 z + a^2} 4 \sqrt z \sqrt {a^2 - z} + \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {a^2 - z} } } + C\) Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$
\(\ds \) \(=\) \(\ds \frac {-2 z + a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^4} {16} \paren {2 \sinh^{-1} \sqrt {\frac z {a^2} } } + C\) Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\ds \) \(=\) \(\ds \frac {-2 z + 2 a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^4} 8 \sinh^{-1} \sqrt {\frac z {a^2} } + C\)
\(\ds \) \(=\) \(\ds \frac {a^2 - z} 4 \sqrt z \sqrt {a^2 - z} + \frac {a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^4} 8 \sinh^{-1} \frac {\sqrt z} a + C\)
\(\ds \) \(=\) \(\ds \frac {-x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 + \frac {a^4} 8 \sinh^{-1} \frac x a + C\) substituting for $z$

$\blacksquare$


Also see


Sources

(in which a mistake apppears)
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