\section{Primitive of x by Arccosine of x over a}
Tags: Primitives involving Inverse Cosine Function, Primitive of x by Arccosine of x over a

\begin{theorem}
:$\displaystyle \int x \arccos \frac x a \ \mathrm d x = \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + C$
\end{theorem}

\begin{proof}
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccos \frac x a
      | c = 
}}
{{eqn | n = 1
      | ll= \implies
      | l = \cos u 
      | r = \frac x a
      | c = Definition of [[Definition:Arccosine|Arccosine]]
}}
{{eqn | n = 2
      | ll= \implies
      | l = \sin u 
      | r = \sqrt {1 - \frac {x^2} {a^2} }
      | c = [[Sum of Squares of Sine and Cosine]]
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arccos \frac x a \ \mathrm d x
      | r = -a \int u \left({a \cos u}\right) \sin u \ \mathrm d u
      | c = [[Primitive of Function of Arcsine]]
}}
{{eqn | r = -a \int a u \frac {\sin 2 u} 2 \ \mathrm d u
      | c = [[Double Angle Formula for Sine]]
}}
{{eqn | r = -\frac {a^2} 2 \int u \sin 2 u \ \mathrm d u
      | c = [[Primitive of Constant Multiple of Function]]
}}
{{eqn | r = -\frac {a^2} 2 \left({\frac {\sin 2 u} 4 - \frac {u \cos 2 u} 2}\right) + C
      | c = [[Primitive of x by Sine of a x|Primitive of $x \sin a x$]] where $a = 2$
}}
{{eqn | r = -\frac {a^2 \sin 2 u} 8 + \frac {a^2 u \cos 2 u} 4 + C
      | c = simplifying
}}
{{eqn | r = -\frac {a^2 \sin u \cos u} 4 + \frac {a^2 u \cos 2 u} 4 + C
      | c = [[Double Angle Formula for Sine]]
}}
{{eqn | r = -\frac {a^2 \sin u \cos u} 4 + \frac {a^2 u \left({\cos^2 u - \sin^2 u}\right)} 4 + C
      | c = [[Double Angle Formula for Cosine]]
}}
{{eqn | r = -\frac {a^2 \sin u \cos u} 4 + \frac {a^2 \arccos \frac x a \left({\cos^2 u - \sin^2 u}\right)} 4 + C
      | c = substituting for $u$
}}
{{eqn | r = -\frac {a^2 \sin u \frac x a} 4 + \frac {a^2 \arccos \frac x a \left({\frac {x^2} {a^2} - \sin^2 u}\right)} 4 + C
      | c = substituting for $\cos u$ from $(1)$
}}
{{eqn | r = -\frac {a^2 \sqrt {1 - \frac {x^2} {a^2} } \frac x a} 4 + \frac {a^2 \arccos \frac x a \left({\frac {x^2} {a^2} - \left({1 - \frac {x^2} {a^2} }\right)}\right)} 4 + C
      | c = substituting for $\sin u$ from $(2)$
}}
{{eqn | r = \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
\end{proof}

\begin{proof}
With a view to expressing the [[Definition:Primitive|primitive]] in the form:
:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccos \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-1} {\sqrt {a^2 - x^2} }
      | c = [[Derivative of Arccosine of x over a|Derivative of $\arccos \dfrac x a$]]
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = [[Primitive of Power]]
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arccos \frac x a \rd x
      | r = \frac {x^2} 2 \arccos \frac x a - \int \frac {x^2} 2 \paren {\frac {-1} {\sqrt {a^2 - x^2} } } \rd x + C
      | c = [[Integration by Parts]]
}}
{{eqn | r = \frac {x^2} 2 \arccos \frac x a + \frac 1 2 \int \frac {x^2 \rd x} {\sqrt {a^2 - x^2} } + C
      | c = [[Primitive of Constant Multiple of Function]]
}}
{{eqn | r = \frac {x^2} 2 \arccos \frac x a + \frac 1 2 \paren {\frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} + C
      | c = [[Primitive of x squared over Root of a squared minus x squared|Primitive of $\dfrac {x^2} {\sqrt {a^2 - x^2} }$]]
}}
{{eqn | r = \frac {x^2} 2 \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + \frac {a^2} 4 \paren {\frac \pi 2 - \arccos \frac x a} + C
      | c = [[Sum of Arcsine and Arccosine]]
}}
{{eqn | r = \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + \frac {\pi a^2} 8 + C
      | c = simplifying
}}
{{eqn | r = \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + C
      | c = subsuming $\dfrac {\pi a^2} 8$ into the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]
}}
{{end-eqn}}
{{qed}}
\end{proof}


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