Primitive of Square of Arccosine of x over a

Theorem

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

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

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

\(\ds \leadsto \ \ \)

\(\ds \sin u\)

\(=\)

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

Sum of Squares of Sine and Cosine

Then:

\(\ds \int \paren {\arccos \frac x a}^2 \rd x\)

\(=\)

\(\ds -a \int u^2 \sin u \rd x\)

Primitive of Function of Arccosine

\(\ds \)

\(=\)

\(\ds -a \paren {2 u \sin u + \paren {2 - u^2} \cos u} + C\)

Primitive of $x^2 \sin a x$ where $a = 1$

\(\ds \)

\(=\)

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

substituting for $u$, $\sin u$ and $\cos u$

\(\ds \)

\(=\)

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

simplifying

$\blacksquare$

Also see

• Primitive of $\paren {\arcsin \dfrac x a}^2$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.482$