Primitive of Arccosine Function/Proof 2

Theorem

$\ds \int \arccos x \rd x = x \arccos x - \sqrt {1 - x^2} + C$

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

$\ds u$

$=$

$\ds \arccos x$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac {-1} {\sqrt {a^2 - 1} }$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Then:

$\ds \int \arccos x \rd x$

$=$

$\ds x \arccos x - \int x \paren {\frac {-1} {\sqrt {1 - x^2} } } \rd x + C$

$\ds $

$=$

$\ds x \arccos x + \int \frac {x \rd x} {\sqrt {1 - x^2} } + C$

simplifying

$\ds $

$=$

$\ds x \arccos x + \paren {-\sqrt {1 - x^2} } + C$

$\ds $

$=$

$\ds x \arccos x - \sqrt {1 - x^2} + C$

simplifying

$\blacksquare$