<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] I'll try to convey general lessons by colouring some coefficients. The crux of this problem is to change the square-rooted expression $4x-x^2=\color{blue}{2}^2-(x-\color{limegreen}{2})^2$ to a squared trigonometric function with $x=\color{limegreen}{2}+\color{blue}{2}\sin t=2(1+\sin t)$, so the integral becomes$$\begin{align}\int\frac{-7x^2dx}{\sqrt{4-(2-x)^2}}&=-28\int(1+\sin t)^2dt\\&=14\int(-3-4\sin t+\cos 2t)dt\\&=14(-3t+4\cos t\color{red}{+}\tfrac12\sin 2t)+C\\&=14(-3\arcsin\tfrac{x-2}{2}\color{red}{+}2\sqrt{4x-x^2}+\tfrac{x-2}{\color{red}{4}}\sqrt{4x-x^2})+C,\end{align}$$where in the last line we use$$\sin t=\frac{x-2}{2},\,\cos t=\sqrt{1-\left(\frac{x-2}{2}\right)^2}=\frac12\sqrt{4x-x^2},\,\sin 2t=2\sin t\cos t.$$Of course, you already knew most of this. But everything I've marked in red corrects a sign error in your result (plus a coefficient error @Jam noted), probably due to forgetting that $\cos^\prime u=\color{red}{-}\sin u$ but $\int\cos u=\color{red}{+}\sin u+C$.
</s>