Primitive of Reciprocal of Root of a squared minus x squared cubed

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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 1

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\paren {\sqrt {a^2 - x^2} }^3}\) \(=\) \(\ds \int \frac {\d z} {2 \sqrt z \paren {\sqrt {a^2 - z} }^3}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} {\paren {a^2 - z} \sqrt z \sqrt {a^2 - z} }\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 \sqrt z} {a^2 \sqrt {a^2 - z} } } + C\) Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 x} {a^2 \sqrt {a^2 - x^2} } } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac x {a^2 \sqrt {a^2 - x^2} } + C\) simplifying

$\blacksquare$


Proof 2

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 $a \cos u$

$\blacksquare$


Also see


Sources

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