Primitive of x by Root of x squared minus a squared cubed
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Theorem
- $\ds \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + C$
for $\size x \ge a$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2 - a^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x\) | \(=\) | \(\ds \int \frac {z^{3/2} } 2 \rd z\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \frac {z^{5/2} } {\frac 5 2} + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {z^{5/2} } 5 + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + C\) | substituting for $z$ |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.231$