Primitive of Arccosine of x over a/Proof 1

Theorem

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

Proof

Let:

\(\ds u\)

\(=\)

\(\ds \arccos \frac x a\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \cos u\)

\(=\)

\(\ds \frac x a\)

Definition of Real Arccosine

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \cos u\)

\(=\)

\(\ds \sqrt {1 - \frac {x^2} {a^2} }\)

Sum of Squares of Sine and Cosine

Then:

\(\ds \int \arccos \frac x a \rd x\)

\(=\)

\(\ds -a \int u \sin u \rd u\)

Primitive of Function of Arccosine

\(\ds \)

\(=\)

\(\ds -a \paren {\sin u - u \cos u} + C\)

Primitive of $x \sin a x$

\(\ds \)

\(=\)

\(\ds -a \paren {\sin u - u \frac x a} + C\)

Substitution for $\cos u$ from $(1)$

\(\ds \)

\(=\)

\(\ds -a \paren {\sqrt {1 - \frac {x^2} {a^2} } - u \frac x a} + C\)

Substitution for $\sin u$ from $(2)$

\(\ds \)

\(=\)

\(\ds -a \paren {\sqrt {1 - \frac {x^2} {a^2} } - \frac x a \arccos \frac x a} + C\)

Substitution for $u$

\(\ds \)

\(=\)

\(\ds x \arccos \frac x a - \sqrt {a^2 - x^2} + C\)

simplifying

$\blacksquare$