Primitive of Reciprocal of Root of a squared minus x squared cubed/Proof 2
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Theorem
- $\ds \int \frac {\d x} {\paren {\sqrt {a^2 - x^2} }^3} = \frac x {a^2 \sqrt {a^2 - x^2} } + C$
Proof
Let:
\(\ds x\) | \(=\) | \(\ds a \sin u\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d x} {\d u}\) | \(=\) | \(\ds a \cos u\) | Derivative of Sine Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2 - x^2\) | \(=\) | \(\ds a^2 \cos^2 u\) | Sum of Squares of Sine and Cosine |
Hence:
\(\ds \int \frac {\d x} {\paren {\sqrt {a^2 - x^2} }^3}\) | \(=\) | \(\ds \int \frac {a \cos u \rd u} {a^3 \cos^3 u}\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \int \sec^2 u \rd u\) | Secant is Reciprocal of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \tan u + c\) | Primitive of Secant Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \dfrac {a \sin u} {a \cos u} + C\) | Definition of Real Tangent Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac x {a^2 \sqrt {a^2 - x^2} } + C\) | substituting for $a \sin u$ and $a \cos u$ |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration: Useful substitutions: Example