Having trouble with this one.
$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}dx$
Note that $\displaystyle 4x - x^2 = 4 - (x + 2)^2$.
Make the substitution $\displaystyle x + 2 = 2 \sin t$.
You might want to check the details (as I rushed this) - I get something like $\displaystyle 8 \int (\sin t - 1)^2 \, dt$.
Expand, integrate term-by-term and finally substitute back that $\displaystyle x + 2 = 2 \sin t$ (tedious, I know, but not too hard).
Ask for more help if you get stuck at any point.
Edit: There is a small mistake - you should spot it easily. A small detail that doesn't change the general game plan .....
$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx$
Let $\displaystyle x-2=2\sin u$ then,
$\displaystyle \int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2 u+2\sin u+1\:du$, now you integrate each term
now you should get, $\displaystyle 4\int \sin^2 u+2\sin u+1\:du=6u-2\sin u\cos u-8\cos u$
Re-substitute, each term and your final answer should/will be:
$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6\arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)\sqrt{4x-x^2}}{2}+C$
I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get $\displaystyle 6\theta-8\cos\:\theta-\sin 2\theta$.
It appears that polymerase used the double-angle formula to get $\displaystyle -2\sin u\cos u$ instead of $\displaystyle -\sin 2 u$? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around.