How to find the indefinite integral? $$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
This is the farthest I've got:
$$=\int\frac{x^2}{\sqrt{1-(x-1)^2}}dx$$
 A: Hint: substitute $$x-1= \sin t $$ so as $$\\ x-1= \sin t \\ x=\sin t +1\\ dx = \cos t\,dt \\ \int  \frac { x^2 }{ \sqrt { 1-(x-1)^2 }  } \, dx=\int  \frac { \cos t (\sin t +1)^2 }{ \sqrt { 1-\sin^2{t} }  } \, dt=\int (\sin t +1)^2 \,  dt \\ $$
A: As $0<x<2,$
$$\dfrac{x^2}{\sqrt{2x-x^2}}=\dfrac{x^{3/2}}{\sqrt{2-x}}$$
set $x=2\sin^2t,x^{3/2}=\text{?}$
$dx=\text{?}$ and $\sqrt{2-x}=+\sqrt2\cos t$
A: Hint:
As $\dfrac{d(2x-x^2)}{dx}=2-2x$
$$\dfrac{x^2}{\sqrt{2x-x^2}}=\dfrac{x^2-2x+2x-2+2}{\sqrt{2x-x^2}}$$
$$=-\sqrt{1-(x-1)^2}-\dfrac{2-2x}{\sqrt{2x-x^2}}+\dfrac2{\sqrt{1-(x-1)^2}}$$
Now use $\#1,\#8$ of this
A: Ok, so building off of what lab bhattacharjee said:
$$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
$$=-\int\sqrt{1-(x-1)^2}dx-\int\dfrac{2-2x}{\sqrt{2x-x^2}}dx+2\int\dfrac1{\sqrt{1-(x-1)^2}}dx$$
Ok, so I use #8 on the 1st integral, u-substitution on the 2nd, and #1 on the 3rd.
$$=-(\frac{(x-1)\sqrt{1-(x-1)^2}}{2}+\frac{1}{2}\arcsin(x-1))-2\sqrt{2x-x^2}+2\arcsin(x-1)+C$$
Simplify.
$$=-\frac{(x-1)\sqrt{1-(x-1)^2}}{2}-2\sqrt{2x-x^2}+\frac{3}{2}\arcsin(x-1)+C$$
