Primitive of Arcsine of x over a/Proof 1

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Theorem

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


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:

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

$\blacksquare$