Primitive of Arcsine of x over a

Theorem

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

Proof 1

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

$\blacksquare$

Proof 2

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\ds u$ $=$ $\ds \arcsin \frac x a$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds \frac 1 {\sqrt {a^2 - x^2} }$ Derivative of $\arcsin \dfrac x a$

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds x$ Primitive of Constant

Then:

 $\ds \int \arcsin \frac x a \rd x$ $=$ $\ds x \arcsin \frac x a - \int x \paren {\frac 1 {\sqrt {a^2 - x^2} } } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds x \arcsin \frac x a - \int \frac {x \rd x} {\sqrt {a^2 - x^2} } + C$ simplifying $\ds$ $=$ $\ds x \arcsin \frac x a - \paren {-\sqrt {a^2 - x^2} } + C$ Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$ $\ds$ $=$ $\ds x \arcsin \frac x a + \sqrt {a^2 - x^2} + C$ simplifying

$\blacksquare$