Primitive of x squared over Root of a squared minus x squared
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Theorem
\(\ds \int \frac {x^2 \rd x} {\sqrt {a^2 - x^2} }\) | \(=\) | \(\ds \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^2} 2 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} } 2 + C\) | either way round, whichever you prefer |
Proof
With a view to expressing the problem in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds 1\) | Power Rule for Derivatives |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \frac x {\sqrt {a^2 - x^2} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds -\sqrt {a^2 - x^2}\) | Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$ |
Then:
\(\ds \int \frac {x^2 \rd x} {\sqrt {a^2 - x^2} }\) | \(=\) | \(\ds \int x \frac {x \rd x} {\sqrt {a^2 - x^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -x \sqrt {a^2 - x^2} - \int \paren {-\sqrt {a^2 - x^2} } \rd x\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds -x \sqrt {a^2 - x^2} + \int \paren {\sqrt {a^2 - x^2} } \rd x\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds -x \sqrt {a^2 - x^2} + \paren {\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} + C\) | Primitive of $\sqrt {a^2 - x^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C\) |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.239$
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $33$.