Primitive of x cubed by Root of x squared minus a squared
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Theorem
- $\ds \int x^3 \sqrt {x^2 - a^2} \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C$
for $\size x \ge a$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int x^3 \sqrt {x^2 - a^2} \rd x\) | \(=\) | \(\ds \int \frac {z^{3/2} \sqrt {z - a^2} \rd z} {2 \sqrt z}\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \int z \sqrt {z - a^2} \rd z\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac {2 \paren {3 z + 2 a^2} } {15} \paren {\sqrt {z - a^2} }^3 } + C\) | Primitive of $x \sqrt {\paren {a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 z + 2 a^2} {15} \paren {\sqrt {z - a^2} }^3 + C\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 x^2 + 2 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C\) | substituting for $z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 x^2 - 3 a^2 + 5 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 \paren {x^2 - a^2} \paren {\sqrt {a^2 - x^2} }^3 + 5 a^2 \paren {\sqrt {a^2 - x^2} }^3} {15} + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C\) | simplification |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.219$