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

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Theorem

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


Proof

Let:

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

$\blacksquare$



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