Primitive of Root of a squared minus x squared over x

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

Let:

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

$\blacksquare$


Also presented as

This result is also seen presented in the form:

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


Also see


Sources