<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] 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.
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