Primitive of Square of Arcsine of x over a

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

Let:

\(\ds u\) \(=\) \(\ds \arcsin \frac x a\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sin u\) \(=\) \(\ds \frac x a\) Definition of Real Arcsine
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \cos u\) \(=\) \(\ds \sqrt {1 - \frac {x^2} {a^2} }\) Sum of Squares of Sine and Cosine


Then:

\(\ds \int \paren {\arcsin \frac x a}^2 \rd x\) \(=\) \(\ds a \int u^2 \cos u \rd x\) Primitive of Function of Arcsine
\(\ds \) \(=\) \(\ds a \paren {2 u \cos u + \paren {u^2 - 2} \sin u} + C\) Primitive of $x^2 \cos a x$
\(\ds \) \(=\) \(\ds a \paren {2 \arcsin \frac x a \sqrt {1 - \frac {x^2} {a^2} } + \paren {\paren {\arcsin \frac x a}^2 - 2} \frac x a} + C\) substituting for $u$, $\sin u$ and $\cos u$
\(\ds \) \(=\) \(\ds x \paren {\arcsin \frac x a}^2 - 2 x + 2 \sqrt{a^2 - x^2} \arcsin \frac x a + C\) simplifying

$\blacksquare$


Also see


Sources