Primitive of Root of a squared minus x squared cubed
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Theorem
- $\ds \int \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a + C$
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Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \paren {\sqrt {a^2 - x^2} }^3 \rd x\) | \(=\) | \(\ds \int \frac {\paren {\sqrt {a^2 - z} }^3} {2 \sqrt z} \rd x\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt z \paren {\sqrt {a^2 - z} }^3} 4 + \frac {3 a^2} 8 \int \frac {\sqrt {a^2 - z} } {\sqrt z} \rd x + C\) | Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt z \paren {\sqrt {a^2 - z} }^3} 4 + \frac {3 a^2} 8 \paren {\sqrt z \sqrt {a^2 - z} + \frac {a^2} 2 \int \frac {\d x} {\sqrt z \sqrt {a^2 - z} } } + C\) | Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2} 8 x \sqrt {a^2 - x^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {a^2 - x^2} } + C\) | substituting for $z$ and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a + C\) | Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.258$