Primitive of x squared by Root of x squared minus a squared cubed
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Theorem
- $\ds \int x^2 \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^5} 6 + \frac {a^2 x \paren {\sqrt {x^2 - a^2} }^3} {24} - \frac {a^4 x \sqrt {x^2 - a^2} } {16} + \frac {a^6} {16} \ln \size {x + \sqrt {x^2 - a^2} } + C$
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 | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int x^2 \paren {\sqrt {x^2 - a^2} }^3 \rd x\) | \(=\) | \(\ds \int \frac {\sqrt z \paren {z - a^2}^{3/2} } 2 \rd z\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt z \paren {z - a^2}^{5/2} } 6 + \frac {a^2} {12} \int \frac {\paren {z - a^2}^{3/2} } {\sqrt z} \rd z + C\) | Primitive of $\paren {p x + q}^n \sqrt{a x + b}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \paren {x^2 - a^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {x^2 - a^2}^{3/2} \rd x + C\) | substituting for $z$ and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \paren {x^2 - a^2}^{5/2} } 6\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } } + C\) | Primitive of $\paren {\sqrt {x^2 - a^2} }^3$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \paren {x^2 - a^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {x^2 - a^2} }^3} {24}\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac {a^4 x \sqrt {x^2 - a^2} } {16} + \frac {a^6} {16} \ln \size {x + \sqrt {x^2 - a^2} } + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.232$