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

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \frac {\d x} {x^2 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C$


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x^2 \sqrt {a^2 - x^2} }\) \(=\) \(\ds \int \frac {\d z} {2 z \sqrt z \sqrt {a^2 - z} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {a^2 - z} }\)


Using Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$:

$\ds \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$


Setting:

\(\ds x\) \(:=\) \(\ds -z\)
\(\ds m\) \(:=\) \(\ds \frac 3 2\)
\(\ds a\) \(:=\) \(\ds 1\)
\(\ds b\) \(:=\) \(\ds a^2\)


\(\ds \frac 1 2 \int \frac {\d z} {\paren {-z}^{3/2} \sqrt {-z + a^2} }\) \(=\) \(\ds \frac {-\sqrt {-z + a^2} } {2 \paren {\paren {\frac 3 2} - 1} a^2 \paren {-z}^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} a^2} \int \frac {\d z} {\paren {-z}^{\paren {\frac 3 2} - 1} \sqrt {-z + a^2} } + C\) Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a^2 - z} } {a^2 z^{1/2} } - 0 + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C\) substituting back for $z$

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_x_squared_by_Root_of_a_squared_minus_x_squared&oldid=503375"