Primitive of Function of Root of x squared minus a squared

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Theorem

$\ds \int \map F {\sqrt {x^2 - a^2} } \rd x = a \int \map F {a \tan u} \sec u \tan u \rd u$

where $x = a \sec u$.


Proof

First note that:

\(\ds x\) \(=\) \(\ds a \sec u\)
\(\ds \leadsto \ \ \) \(\ds \sqrt {x^2 - a^2}\) \(=\) \(\ds \sqrt {\left({a \sec u}\right)^2 - a^2}\)
\(\ds \) \(=\) \(\ds a \sqrt {\sec^2 u - 1}\) taking $a$ outside the square root
\(\ds \) \(=\) \(\ds a \sqrt {\tan^2 u}\) Difference of Squares of Secant and Tangent
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds a \tan u\)


Then:

\(\ds x\) \(=\) \(\ds a \sec u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d u}\) \(=\) \(\ds a \sec u \tan u\) Derivative of Secant Function
\(\ds \leadsto \ \ \) \(\ds \int \map F {\sqrt {x^2 - a^2} } \rd x\) \(=\) \(\ds \int a \map F {\sqrt {x^2 - a^2} } \sec u \tan u \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int a \map F {a \tan u} \sec u \tan u \rd u\) from $(1)$
\(\ds \) \(=\) \(\ds a \int \map F {a \tan u} \sec u \tan u \rd u\) Primitive of Constant Multiple of Function

$\blacksquare$


Sources