Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form

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Theorem

$\displaystyle \int \frac {\mathrm d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \operatorname{sech}^{-1} {\frac x a} + C$


Proof

Let:

\(\displaystyle u\) \(=\) \(\displaystyle \operatorname{sech}^{-1} {\frac x a}\)
\(\displaystyle x\) \(=\) \(\displaystyle a \operatorname{sech} u\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d x} {\mathrm d u}\) \(=\) \(\displaystyle -a \operatorname{sech} u \tanh u\) Derivative of Hyperbolic Secant Function
\(\displaystyle \implies \ \ \) \(\displaystyle \int \frac {\mathrm d x} {x \sqrt {a^2 - x^2} }\) \(=\) \(\displaystyle \int \frac {-a \operatorname{sech} u \tanh u} {a \operatorname{sech} u \sqrt {a^2 - a^2 \operatorname{sech}^2 u} } \ \mathrm d u\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle -\frac a {a^2} \int \frac {\operatorname{sech} u \tanh u} {\operatorname{sech} u \sqrt {1 - \operatorname{sech}^2 u} } \ \mathrm d u\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 a \int \frac {\operatorname{sech} u \tanh u} {\operatorname{sech} u \tanh u} \ \mathrm d u\) Sum of Squares of Hyperbolic Secant and Tangent
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 a \int 1 \ \mathrm d u\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 a u + C\) Integral of Constant
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 a \operatorname{sech}^{-1} {\frac x a} + C\) Definition of $u$

$\blacksquare$


Also see