Primitive of Function of Root of a squared plus x squared

From ProofWiki
Jump to navigation Jump to search

Theorem

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

where $x = a \tan u$.


Proof

First note that:

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


Then:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Function_of_Root_of_a_squared_plus_x_squared&oldid=503258"