Primitive of x squared by Root of a squared minus x squared/Also presented as
Jump to navigation
Jump to search
Primitive of $x^2 \sqrt {a^2 - x^2}$: Also presented as
This result is also seen presented in the form:
- $\ds \int x^2 \sqrt {a^2 - x^2} \rd x = \frac {a^4} 8 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 x^2} } 8 + C$
Proof
\(\ds \int x^2 \sqrt {a^2 - x^2} \rd x\) | \(=\) | \(\ds -\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \arcsin \frac x a + C\) | Primitive of $x^2 \sqrt {a^2 - x^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \arcsin \frac x a + \frac {a^2 x \sqrt {a^2 - x^2} } 8 - \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + C\) | reversing the order for convenience | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \arcsin \frac x a + \frac {a^2 x \sqrt {a^2 - x^2} - 2 x \paren {\sqrt {a^2 - x^2} }^3} 8 + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 \paren {\sqrt {a^2 - x^2} }^2} } 8 + C\) | Distributive Laws of Arithmetic | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 \paren {a^2 - x^2} } } 8 + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 x^2} } 8 + C\) | arranging into required form |
$\blacksquare$
Sources
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $30$.