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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.246$
- (in which a mistake apppears)