Primitive of x by Root of a squared minus x squared
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Theorem
- $\ds \int x \sqrt {a^2 - x^2} \rd x = \frac {-\paren {\sqrt {a^2 - x^2} }^3} 3 + C$
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int x \sqrt {a^2 - x^2} \rd x\) | \(=\) | \(\ds \int \frac {\sqrt z \sqrt {a^2 - z} \rd z} {2 \sqrt z}\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \int \sqrt {a^2 - z} \rd z\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \frac {2 \paren {\sqrt {a^2 - z} }^3} {-3} + C\) | Primitive of $\sqrt {a x + b}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\paren {\sqrt {a^2 - x^2} }^3} 3 + C\) | substituting for $z$ and simplifying |
$\blacksquare$
Also see
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $8$.
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.245$