Primitive of Function of Root of a squared minus x squared
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Theorem
- $\ds \int \map F {\sqrt {a^2 - x^2} } \rd x = a \int \map F {a \cos u} \cos u \rd u$
where $x = a \sin u$.
Proof
First note that:
\(\ds x\) | \(=\) | \(\ds a \sin u\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt {a^2 - x^2}\) | \(=\) | \(\ds \sqrt {a^2 - \paren {a \sin u}^2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds a \sqrt {1 - \sin^2 u}\) | taking $a$ outside the square root | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds a \cos u\) | Sum of Squares of Sine and Cosine |
Then:
\(\ds x\) | \(=\) | \(\ds a \sin u\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d u}\) | \(=\) | \(\ds a \cos u\) | Derivative of Sine Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \map F {\sqrt {a^2 - x^2} } \rd x\) | \(=\) | \(\ds \int a \map F {\sqrt {a^2 - x^2} } \cos u \rd u\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \int a \map F {a \cos u} \cos u \rd u\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a \int \map F {a \cos u} \cos u \rd u\) | Primitive of Constant Multiple of Function |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Important Transformations: $14.52$