\section{Primitive of Function of Root of a squared minus x squared}
Tags: Integral Substitutions

\begin{theorem}
:$\ds \int \map F {\sqrt {a^2 - x^2} } \rd x = a \int \map F {a \cos u} \cos u \rd u$
where $x = a \sin u$.
\end{theorem}

\begin{proof}
First note that:
{{begin-eqn}}
{{eqn | l = x 
      | r = a \sin u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {a^2 - x^2}
      | r = \sqrt {a^2 - \paren {a \sin u}^2}
      | c =
}}
{{eqn | r = a \sqrt {1 - \sin^2 u}
      | c = taking $a$ outside the square root
}}
{{eqn | n = 1
      | r = a \cos u
      | c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = x 
      | r = a \sin u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = a \cos u
      | c = Derivative of Sine Function
}}
{{eqn | ll= \leadsto
      | l = \int \map F {\sqrt {a^2 - x^2} } \rd x
      | r = \int a \map F {\sqrt {a^2 - x^2} } \cos u \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \int a \map F {a \cos u} \cos u \rd u
      | c = from $(1)$
}}
{{eqn | r = a \int \map F {a \cos u} \cos u \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
\end{proof}

