Primitive of Arccosine Function/Proof 1
Theorem
- $\ds \int \arccos x \rd x = x \arccos x - \sqrt {1 - x^2} + C$
Proof
Let:
| \(\ds u\) | \(=\) | \(\ds \arccos x\) | ||||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \cos u\) | \(=\) | \(\ds x\) | Definition of Real Arccosine | |||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \cos u\) | \(=\) | \(\ds \sqrt {1 - x^2}\) | Sum of Squares of Sine and Cosine |
Then:
| \(\ds \int \arccos x \rd x\) | \(=\) | \(\ds -\int u \sin u \rd u\) | Primitive of Function of Arccosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\sin u - u \cos u} + C\) | Primitive of $x \sin a x$, setting $a := 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\sin u - u x} + C\) | Substitution for $\cos u$ from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\sqrt {1 - x^2} - u x} + C\) | Substitution for $\sin u$ from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\sqrt {1 - x^2} - x \arccos x} + C\) | Substitution for $u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x \arccos x - \sqrt {1 - x^2} + C\) | simplifying |
$\blacksquare$