Primitive of Power of x by Arcsine of x over a/Proof 1

Theorem

$\ds \int x^m \arcsin \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$

Proof

Recall:

$\text {(1)}: \quad$

$\ds \int x^m \arcsin x \rd x$

$=$

$\ds \frac {x^{m + 1} } {m + 1} \arcsin x - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {1 - x^2} }$

Primitive of $x^m \arcsin x$

Then:

$\ds \int x^m \arcsin \frac x a \rd x$

$=$

$\ds \int a^m \paren {\dfrac x a}^m \arcsin \frac x a \rd x$

manipulating into appropriate form

$\ds $

$=$

$\ds a^m \int \paren {\dfrac x a}^m \arcsin \frac x a \rd x$

Primitive of Constant Multiple of Function

$\ds $

$=$

$\ds a^m \paren {\dfrac 1 {1 / a} \paren {\frac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \paren {\dfrac x a}^{m + 1} \frac {\d x} {\sqrt {1 - \paren {\dfrac x a}^2} } } }$

Primitive of Function of Constant Multiple, from $(1)$

$\ds $

$=$

$\ds \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$

simplifying

$\blacksquare$