# Primitive of x by Arcsine of x over a

## Theorem

$\displaystyle \int x \arcsin \frac x a \ \mathrm d x = \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} } 4 + C$

## Proof 1

Let:

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

Then:

 $\displaystyle \int x \arcsin \frac x a \ \mathrm d x$ $=$ $\displaystyle a \int u \left({a \sin u}\right) \cos u \ \mathrm d u$ Primitive of Function of Arcsine $\displaystyle$ $=$ $\displaystyle a \int a u \frac {\sin 2 u} 2 \ \mathrm d u$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle \frac {a^2} 2 \int u \sin 2 u \ \mathrm d u$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {a^2} 2 \left({\frac {\sin 2 u} 4 - \frac {u \cos 2 u} 2}\right) + C$ Primitive of $x \sin a x$ where $a = 2$ $\displaystyle$ $=$ $\displaystyle \frac {a^2 \sin 2 u} 8 - \frac {a^2 u \cos 2 u} 4 + C$ simplifying $\displaystyle$ $=$ $\displaystyle \frac {a^2 \sin u \cos u} 4 - \frac {a^2 u \cos 2 u} 4 + C$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle \frac {a^2 \sin u \cos u} 4 - \frac {a^2 u \left({\cos^2 u - \sin^2 u}\right)} 4 + C$ Double Angle Formula for Cosine $\displaystyle$ $=$ $\displaystyle \frac {a^2 \sin u \cos u} 4 - \frac {a^2 \arcsin \frac x a \left({\cos^2 u - \sin^2 u}\right)} 4 + C$ substituting for $u$ $\displaystyle$ $=$ $\displaystyle \frac {a^2 \frac x a \cos u} 4 - \frac {a^2 \arcsin \frac x a \left({\cos^2 u - \frac {x^2} {a^2} }\right)} 4 + C$ substituting for $\sin u$ from $(1)$ $\displaystyle$ $=$ $\displaystyle \frac {a^2 \frac x a \sqrt {1 - \frac {x^2} {a^2} } } 4 - \frac {a^2 \arcsin \frac x a \left({\left({1 - \frac {x^2} {a^2} }\right) - \frac {x^2} {a^2} }\right)} 4 + C$ substituting for $\cos u$ from $(2)$ $\displaystyle$ $=$ $\displaystyle \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} } 4 + C$ simplifying

$\blacksquare$

## Proof 2

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

 $\displaystyle u$ $=$ $\displaystyle \arcsin \frac x a$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \frac 1 {\sqrt {a^2 - x^2} }$ Derivative of $\arcsin \dfrac x a$

and let:

 $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle x$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {x^2} 2$ Primitive of Power

Then:

 $\displaystyle \int x \arcsin \frac x a \ \mathrm d x$ $=$ $\displaystyle \frac {x^2} 2 \arcsin \frac x a - \int \frac {x^2} 2 \left({\frac 1 {\sqrt {a^2 - x^2} } }\right) \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 \arcsin \frac x a - \frac 1 2 \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} } + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 \arcsin \frac x a - \frac 1 2 \left({\frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a}\right) + C$ Primitive of $\dfrac {x^2} {\sqrt {a^2 - x^2} }$ $\displaystyle$ $=$ $\displaystyle \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} } 4 + C$ simplifying

$\blacksquare$