Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form

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Theorem

For $a > 0$ and $0 < \size x < a$:

$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$


Proof

\(\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} }\) \(=\) \(\ds -\frac 1 a \sech^{-1} {\frac {\size x} a} + C\) Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$: $\sech^{-1}$ form
\(\ds \) \(=\) \(\ds -\frac 1 a \map \ln {\frac {1 + \sqrt {1 - \paren {\frac {\size x} a}^2} } {\frac {\size x} a} } + C\) Definition 2 of Inverse Hyperbolic Secant
\(\ds \) \(=\) \(\ds -\frac 1 a \map \ln {\frac {a + a \sqrt {1 - \paren {\frac {\size x} a}^2} } {\size x} } + C\) multiplying top and bottom by $a$
\(\ds \) \(=\) \(\ds -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - a^2 \paren {\frac {\size x} a}^2} } {\size x} } + C\) moving $a$ within the square root
\(\ds \) \(=\) \(\ds -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C\) simplifying, noting $\size x^2 = x^2$

$\blacksquare$


Also presented as

This result is also seen presented in the form:

$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \ln \size {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$


Also see


Sources

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