Primitive of x over Power of a squared minus x squared
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Theorem
- $\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^n} = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} }$
for $x^2 < a^2$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds a^2 - x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds -2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^n}\) | \(=\) | \(\ds \int \frac {\d z} {- 2 z^n}\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} 2 \int z^{-n} \rd z\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} 2 \frac {z^{-n + 1} } {-n + 1} + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \paren {n - 1} z^{n - 1} } + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} } + C\) | substituting for $z$ |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.178$