<s>[INST] Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here. 
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{3/2}}{\sqrt{4-x}} dx \\
&= -7 \int\frac{8\sin^3\theta\  2\cos\theta}
          {\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
2\cos\theta = dx \\
(2\sin\theta)^3 = x^{3/2} 
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\  2\cos\theta}
          {2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -56 \int \sin^3\theta\, d\theta \\
&= -56 \int (1 - \cos^2\theta) \sin\theta\, d\theta \\
&= -56 \int(\sin\theta - \cos^2\theta\sin\theta)\,d\theta \\
&= 56 \cos\theta - 56 \int u^2\, du &&
\begin{array}{c}
u = \cos\theta \\
du = \sin\theta
\end{array}\\
&= 56 \cos\theta - 56 \frac{\cos^3\theta}{3} + C \\
&= 56 \left(\frac{\sqrt{4-x}}{2}
  - \frac13 \left( \frac{\sqrt{4-x}}{2}\right)^3 \right)+ C 
&& \cos\theta = \frac{\sqrt{4-x}}{2}\\
&= 56 \frac{\sqrt{4-x}}{2}
  \left(1 - \frac{4-x}{12}\right) + C\\
\end{align}
My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?
My second question is: is this a correct solution?
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C$$
It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p
Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some specific advice about how I could improve my math tricks to that point?
 [/INST] Disclaimer: Since you are saying that it is your last attempt on Webwork, so I would suggest that you should read through the steps carefully to see if there is any typo or mistake on my part.The approach is along the same lines however.
Since $4x-x^2=-(x^2-4x+2^2-2^2) =-(x-2)^2+2^2 $
So 
$\int \frac{x^2}{\sqrt{4x-x^2}}dx =\int \frac{x^2}{\sqrt{2^2-(x-2)^2}}dx$
Let $x-2 = 2 \sin u$ then $dx = 2 \cos u du$
Now back to integral 
$\int \frac{x^2}{\sqrt{2^2-(x-2)^2}}dx = \int \frac{(2+2\sin u)^2}{\sqrt{2^2-(2\sin u)^2}}2\cos udu$
$= \int \frac{(2+2\sin u)^2}{\sqrt{2^2(1-\sin^2 u)}}2\cos udu$
$= \int \frac{(2+2\sin u)^2}{\sqrt{2^2(\cos^2 u)}}2\cos udu$
$= \int \frac{(2+2\sin u)^2}{2(\cos u)}2\cos udu$
$= \int (2+2\sin u)^2du$
$= 4(\int (1+\sin^2 u+2\sin u ))du$
$= 4(\int 1 du + \int \sin^2 u du +2 \int \sin u) du$
$= 4(u + \int \sin^2 u du -2 \cos u )$
$= 4(u + \int \sin^2 u du -2 \cos u )$
$=4(u + \frac{1}{2}\left(\int (1-\cos 2u) du \right) -2 \cos u) $
$=4( u + \frac{1}{2}\left(u- \frac{\sin 2u}{2}  \right) -2 \cos u )$
$= 4(u + \frac{1}{2}\left(u- \frac{2\sin u\cos u}{2}  \right) -2 \cos u )$
$=4( u + \frac{1}{2}\left(u- {\sin u\cos u}  \right) -2 \cos u )$
$=4( \frac{3u}{2} - \frac{1}{2}{\sin u\cos u}  -2 \cos u )$
$= 6u - 2\sin u\cos u  -8 \cos u $
Since  $\sin u = \frac{x -2}{2}$   So     
$\cos u = \sqrt{1-\sin^2 u} = \sqrt{1-(\frac{x -2}{2})^2} = \sqrt{\frac{4 -x^2 +4x -4}{4}} =\frac{\sqrt{4x -x^2}}{2}$ 
and  
$u = \arcsin(\frac{x -2}{2})$
which leads to 
$\int \frac{x^2}{\sqrt{4x-x^2}}dx = 6 \arcsin(\frac{x -2}{2}) - 2 \frac{x -2}{2}\frac{\sqrt{4x -x^2}}{2}  -8 \frac{\sqrt{4x -x^2}}{2} $
$= 6 \arcsin(\frac{x -2}{2}) -  \frac{(x -2)\sqrt{4x -x^2}}{2}  -4 \sqrt{4x -x^2}$
Now you can combine your original integral and the integral constant $C$ as
$\int \frac{-7x^2}{\sqrt{4x-x^2}}dx = -7\left( 6 \arcsin(\frac{x -2}{2}) -  \frac{(x -2)\sqrt{4x -x^2}}{2}  -4 \sqrt{4x -x^2} \right)+C$
</s>