Help with trig sub integral I've gone over my work twice, and while I dropped a negative the first time, I believe that my work is sound. Webwork is not accepting it, however.
Here is the problem and my solution. Sorry for bad handwriting

 A: Hint:
$$\int\dfrac{x^2}{\sqrt{4x - x^2}}\,\mathrm dx\equiv\int\dfrac{x^2}{\sqrt{4 - (x - 2)^2}}\,\mathrm dx$$
Let $x - 2 = 2\sin(u)\implies\mathrm dx = 2\cos(u)\,\mathrm du$. Therefore,
$$\begin{align}\int\dfrac{x^2}{\sqrt{4 - (x - 2)^2}}\,\mathrm dx&\equiv\int\dfrac{2\cos(u)(2\sin(u) + 2)^2}{\sqrt{4 - 4\sin(u)}}\,\mathrm du \\ &\stackrel{\sin^2(u) + \cos^2(u) = 1}=4\int(\sin(u) + 1)^2\,\mathrm du \\ &= 4\int\sin^2(u) + 2\sin(u) + 1\,\mathrm du \\ &=4\int\sin^2(u)\,\mathrm du + 8\int\sin(u)\,\mathrm du + 4\int1\,\mathrm du\end{align}$$
Can you take it from here?
A: Let $x=2\sin{t}$.
Thus, $$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{x^2-4+4}{\sqrt{4-x^2}}dx=4\int\frac{1}{\sqrt{4-x^2}}dx-\int\sqrt{4-x^2}dx=$$
$$=4\arcsin\frac{x}{2}-4\int\cos^2tdt=4\arcsin\frac{x}{2}-2\int(1+\cos2t)dt.$$
Can you end it now?
Also, let $x-2=v$.
Thus, we can use the similar way: $$\int\frac{x^2}{\sqrt{4x-x^2}}dx=\int\frac{x^2}{\sqrt{4-(x-2)^2}}dx=\int\frac{v^2+4v+4}{\sqrt{4-v^2}}dv=$$
$$=\int\frac{v^2-4+4v+8}{\sqrt{4-v^2}}dv=\int\frac{4v}{\sqrt{4-v^2}}dv-\int\sqrt{4-v^2}dv+8\int\frac{1}{\sqrt{4-v^2}}dv.$$
