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?
 A: 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 $\int\cos u=\color{red}{+}\sin u+C$.
A: 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
$\int \frac{-7x^2}{\sqrt{4x-x^2}}dx = -7\left( 6 \arcsin(\frac{x -2}{2}) -  \frac{(x -2)\sqrt{4x -x^2}}{2}  -4 \sqrt{4x -x^2} \right)+C$
A: You do not have a "$\theta$" on your diagram, so it is not possible to verify that $\theta$ and $x$ have a correct relationship.  You do not write "$x = \dots$", so I have to guess that you mean
$$  x = 4 \sin^2 \theta  \text{.}  $$
The original integrand is only defined on $x \in [0,4]$, so happily this choice of substitution for $x$ is capable of representing the entire valid range of $x$s in the original integral.
From this, 
$$  \mathrm{d}x = 8 \cos \theta \sin \theta \,\mathrm{d}\theta  \text{,}  $$
which is not equivalent to your "$2 \cos \theta = \mathrm{d}x$".
The derivative of your proposed solution with respect to $x$ is
$$  \frac{-7x^2 + 84x - 126}{\sqrt{4x-x^4}}  \text{,}  $$
which is not (up to a constant of integration) equivalent to the given integrand.
Using only algebra, substitution, and trig substitution, I would attack your integral as \begin{align*}
\int & \; \frac{-7 x^2}{\sqrt{4x-x^2}} \,\mathrm{d}x  \\
    &= \int \; \frac{-7 x^2}{\sqrt{-(x^2 - 4x + 4 - 4)}} \,\mathrm{d}x  \\
    &= \int \; \frac{-7 x^2}{\sqrt{-((x-2)^2 - 4)}} \,\mathrm{d}x  &\begin{bmatrix} u = x-2 \\ \mathrm{d}u = \mathrm{d}x \end{bmatrix}  \\
    &= \int \; \frac{-7 (u+2)^2}{\sqrt{4 - u^2}} \,\mathrm{d}u  &\begin{bmatrix} u = 2\cos\theta \\ \mathrm{d}u = -2\sin\theta\,\mathrm{d}\theta\end{bmatrix}  \\
    &= \int \; \frac{-7 (2\cos\theta+2)^2}{\sqrt{4 - 4\cos^2 \theta}} (-2 \sin\theta) \,\mathrm{d}\theta  \\
    &= \int \; \frac{-7 (2\cos\theta+2)^2}{2 \sin \theta} (-2 \sin\theta) \,\mathrm{d}\theta  \\
    &= 7 \int \; (2\cos\theta+2)^2 \,\mathrm{d}\theta  \text{.}
\end{align*}
Then multiply out the binomial and split the sum into three integrals.  The constant integral is easy.  The cosine integral is easy.  The squared cosine integral is easy if you use reduction formulas (the fifth and sixth entries in the table here) or the identity $\cos 2x = 2 \cos^2 x - 1$ to double the angle and lower the power.
A: 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.
