Help with arcsin integral I have the following integral:
$$ I =\int x^2\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) dx$$
Where $a$ and $b$ are non-zero positive integers, and $x<a$. I have started by integration by parts:
$$ I = uv - \int vu'$$
$$u = \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right )$$
$$u' =  \left [ \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) \right ]'$$
We substitute the inside content of the arcsin by $z=\frac{\sqrt{a^2-x^2}}{b}$:
$$u' =  \left [ \sin^{-1}(z) \right ]' = \frac{1}{\sqrt{1-z^2}} \cdot  z'$$
$$z' = - \frac{x}{b} (a^2-x^2)^{- \frac{1}{2}}$$
Hence we obtain:
$$u' =  -\frac{x}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$
On the other hand of the integration by parts we have $v'$:
$$v' = x^2$$
$$v = \frac{1}{3}x^3$$
Hence we end up with:
$$I = \frac{1}{3}x^3\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) + \frac{1}{3}\int \frac{x^4}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$
I cant really figure out how to solve this final integral. Any help would be appreciated

Make the substitution
$$u = \frac{\sqrt{a^2-x^2}}b.$$
Then $x = \sqrt{a^2 - u^2 b^2},$ and so $d x = - \frac{b^2 du}{2 \sqrt{a^2-u^2}},$
and so the integral becomes
$$\frac{b^2}2\int \sqrt{a^2 - u^2 b^2} \arcsin(u) du= \frac{b^2 a}{2} \int \sqrt{1- (\frac{ub}a)^2} \arcsin u du$$ Now integrate by parts.
