Primitive of Power of x by Arcsine of x over a

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Theorem

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


Proof

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle x^m\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1}\) Primitive of Power


Then:

\(\displaystyle \int x^m \arcsin \frac x a \rd x\) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \int \frac {x^{m + 1} } {m + 1} \left({\frac 1 {\sqrt {a^2 - x^2} } }\right) \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt{a^2 - x^2} } + C\) Primitive of Constant Multiple of Function

$\blacksquare$


Also see


Sources