Primitive of Arcsine of x over a/Proof 1
Jump to navigation
Jump to search
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 |
$\blacksquare$