After trigonometric substitution, writing the antiderivative in terms of $x.$ 
The following integral suggests trigonometric substitution $x=4 \sin (\theta)$ :
$$
\int \frac{x^2}{\left(16-x^2\right)^{3 / 2}} d x \text {. }
$$
After making this substitution and integrating, we obtain
$$
\int \dfrac{x^2}{\left(16-x^2\right)^{3 / 2}} d x=\int \dfrac{\sin ^2(\theta) \cos (\theta)}{\left(\cos ^2(\theta)\right)^{3 / 2}} d \theta=\int \tan ^2(\theta) d \theta=\int \sec ^2(\theta)-1 d \theta=\tan (\theta)-\theta+C .
$$
The final step is to return to the original variable $x$. Do NOT use sine, cosine, or tangent
functions in your final answer (but you may have inverse trigonometric functions in your
answer).

Image of the right triangle:

I'm honestly not following much of this and need a good starting point. All I can really deduce is that since the integration amounts to $\tan(\theta) - \theta + C$, then perhaps that's telling me something about the TOA in SOHCAHTOA? That there's a relationship between the angle it's showing highlighted in the diagram and the opposite/adjacent angle? Is that right?
 A: Before answering your question, I will assume that your function that you are integrating is defined on $(-4, 4)$ (the largest domain it can have, assuming it is a real valued function). Therefore, the substitution $x=4\sin({\theta})$ "transforms" your function domain in the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ (this is not the only interval that can be chosen, but this is the easiest one to work with).
Now to answer your question: We are working with the function $g\colon \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \to \mathbb{R}, \;g(\theta)=\tan(\theta)-\theta$, where $x=4\sin(\theta) \iff \sin(\theta)=\frac{x}{4}, x \in (-4, 4)$. Because $\theta$ ranges in $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, we can take the inverse sine on both sides, which results in $\theta=\arcsin\left(\frac{x}{4}\right)$. This argument would not work if $\theta$ was, for example, in $\left( 0, \pi\right)$, because the $\arcsin$ function is defined as follows: $\arcsin\colon[-1, 1] \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.
Now let $$t \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \implies \cos(t) > 0 \implies \sqrt{\cos^2(t)}=\cos(t) \implies \tan(t)=\frac{\sin(t)}{\cos(t)}=\frac{\sin(t)}{\sqrt{1-\sin^2(t)}} \implies \tan(\arcsin(a))=\frac{a}{\sqrt{1-a^2}}, \text{ where } a \in (-1, 1)$$
In your integral problem, you have $$\sin(\theta) = \frac{x}{4}, \; x\in(-4, 4) \implies \tan(\theta)-\theta=\frac{\frac{x}{4}}{\sqrt{1-{\left(\frac{x}{4}\right)}^2}}-\arcsin\left(\frac{x}{4}\right)=\frac{x}{\sqrt{16-x^2}}-\arcsin(\frac{x}{4})$$
As pointed out in the comments, one could deduce this formula from a right-angle triangle, by using the Pythagorean Theorem: 
You can take the angle $\theta$ to be $\angle ABC$ in the image, which would make $AC=x$ and $BC=4$. Now we can deduce that $AB=\sqrt{16-x^2}$, which implies $\tan(\theta)=\frac{AC}{AB}=\frac{x}{\sqrt{16-x^2}}$. This can be easily generalized for angles $\theta \in \left(-\frac{\pi}{2}, 0\right]$ and you can check by using the same method that the formula still holds.
Moreover, if you know that $\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}}$, you can prove it by taking $f\colon [-1, 1] \to \mathbb{R}, \; f(x)=\tan(\arcsin(x))-\frac{x}{\sqrt{1-x^2}}$. You can differentiate the function and see that $\frac{df}{dx}=0, \forall x \in [-1, 1]$, which means $f$ is a constant function, therefore $f(x)=f(0)=0$, which proves the claim.
I recommend that you get familiar with all kinds of computations, such as $\sin(\arctan(x))$ or $t(t^{-1}(x))$, where $t$ is a trigonometric function. This type of computation will show up many times in trigonometric substitutions.
A: Without any substitution...at the beginning:
$$\int x\cdot x(16-x^2)^{-3/2} dx\stackrel{\text{by parts}\,u=x, v'=x(16-x^2)^{-3/2}}=x(16-x^2)^{-1/2}-\int (16-x^2)^{-1/2}dx\;\;(*)$$
Substitution in the integral: $\;\frac x4=\sin t\implies dx=4\cos t\,dt\;$ , so
$$(*)=x(16-x^2)^{-1/2}-\int\left(1-\sin^2t\right)^{-1/2}\cos t\,dt=x(16-x^2)^{-1/2}-t+C=$$
$$=x(16-x^2)^{-1/2}-\arcsin\frac x4+C$$
