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?

The mistake in your original attempt has been identified:
the substitution for $dx.$
In particular, 
$$ \frac{d}{d\theta} 4 \sin^2 \theta = 8 \sin\theta\cos\theta \neq 2 \cos\theta.$$
Patching your original attempt by making the correct substitution,
\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\  8 \sin\theta\cos\theta}
          {\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
8 \sin\theta\cos\theta \,d\theta = dx \\
(2\sin\theta)^3 = x^{3/2} 
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\  8 \sin\theta\cos\theta}
          {2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -224 \int \sin^4\theta\, d\theta \\
&= -224 \int (1 - \cos^2\theta) \sin^2\theta\, d\theta \\
&= -224 \int(\sin^2\theta - \cos^2\theta\sin^2\theta)\,d\theta \\
\end{align}
Here we have to switch strategies due to the extra factor $\sin\theta.$
\begin{align}
\int\sin^2\theta \,d\theta &= \int \frac12 (1 - \cos(2\theta)) \,d\theta\\
&= \frac12\theta - \frac12 \int \cos(2\theta) \,d\theta\\
&= \frac12\theta - \frac14 \sin(2\theta) + C_1
\end{align}
\begin{align}
\int\sin^2\theta\cos^2\theta \,d\theta &= \int \frac14 \sin^2(2\theta) \,d\theta\\
&= \frac18 \int \sin^2 u \,du  && u = 2\theta\\
&= \frac18 \left(\frac12u - \frac14 \sin(2u)\right) + C_2 \\
&= \frac18\theta - \frac1{32} \sin(4\theta) + C_2
\end{align}
So we get that
\begin{align}
-224 \int \sin^4\theta\, d\theta
&=  -224\left(\frac12\theta - \frac14 \sin(2\theta) + C_1
 - \left(\frac18\theta - \frac1{32} \sin(4\theta) + C_2 \right)\right)\\
&=  -224\left(\frac38\theta - \frac14 \sin(2\theta)
        + \frac1{32} \sin(4\theta)\right) + C\\
&=  -7\left(12\theta - 8\sin(2\theta)  + \sin(4\theta)\right) + C\\
&=  -7\left(12\theta - 16\sin\theta\cos\theta
              + 2\sin(2\theta)\cos(2\theta)\right) + C\\
&=  -7\left(12\theta - 16\sin\theta\cos\theta
              + 4\sin\theta\cos\theta(1-2\sin^2\theta)\right) + C\\
&=  -7\left(12\theta - 12\sin\theta\cos\theta
              - 8\sin^3\theta\cos\theta\right) + C\\
&=  -7\left(12\theta - 12\sin\theta\cos\theta
              - 8\sin^3\theta\cos\theta\right) + C\\
&=  -7\left(12\arcsin\left(\frac{\sqrt x}2\right) - 6\sqrt x \sqrt{1 - \frac x4}
              - x^{3/2}\sqrt{1 - \frac x4}\right) + C\\
&=  -7\left(12\arcsin\left(\frac{\sqrt x}2\right) - 3\sqrt{4x - x^2}
              - \frac12 x\sqrt{4x - x^2}\right) + C\\
\end{align}
You might recognize that the terms here are similar to the terms in your proposed solution
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C,$$
especially if you realize that 
$$ \arcsin\left(\frac{\sqrt x}{2}\right)
 = \frac12 \arcsin\left(\frac{x-2}{2}\right) + \frac\pi4, $$
but the coefficients do not match.
The proof of the pudding is to take the derivative.
(I did this for both solutions--and when I first did it on my calculations it showed me I had made an arithmetic error, which I was then able to find and correct to get the solution shown here now.)

By the way, I found this a tedious and error-prone approach. Other solutions offer better approaches. I was just curious to see how your original attempt would work out without mistakes.
