Primitive of x squared by Root of x squared minus a squared/Also presented as
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Primitive of $x^2 \sqrt {x^2 - a^2}$: Also presented as
This result is also seen presented in the form:
- $\ds \int x^2 \sqrt {x^2 - a^2} \rd x = \frac x 8 \paren {2 x^2 - a^2} \sqrt {x^2 - a^2} - \frac {a^4} 8 \arcosh \dfrac x a + C$
Proof
\(\ds \int x^2 \sqrt {x^2 - a^2} \rd x\) | \(=\) | \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C\) | Primitive of $x^2 \sqrt {x^2 - a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \arcosh \dfrac x a + C\) | Definition of Inverse Hyperbolic Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 x \paren {\sqrt {x^2 - a^2} }^3 + a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \arcosh \dfrac x a + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x \sqrt {x^2 - a^2} \paren {2 \paren {x^2 - a^2} + a^2} } 8 - \frac {a^4} 8 \arcosh \dfrac x a + C\) | Distributive Laws of Arithmetic | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac x 8 \paren {2 x^2 - a^2} \sqrt {x^2 - a^2} - \frac {a^4} 8 \arcosh \dfrac x a + C\) | simplifying |
$\blacksquare$
Sources
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $41$.