How do I solve this problem with U-substitution? 
$$\int \left(4-x(16-x^2)^{1/2}\right)\,dx $$

I learned today I could use U-substitution to before integrating, which makes it easier to integrate.
So I can make $U=16-x^2,\quad \dfrac{du}{dx} = -2x^2$
And I've no idea how I should proceed next. 
My teacher only showed me an easy example: 
$$x(x+3)^{1/2} \implies U=x+3,\;\; \frac{du}{dx} = 1$$ 
$$\text{Substitution:}\quad x(x+3)^{1/2} \quad = \quad(U-3)U^{1/2}$$
But I can't figure out on my own how to proceed with the integral on top.
 A: You chose the correct $$U:\;\;\color{red}{\bf U = 16 - x^2}\implies dU = -2x\,dx \implies \color{blue}{\bf x\,dx = -\frac 12\, dU}\tag{1}$$
Substituting terms of $(2)$ into the equivalent substitutions shown in $(1)$ gives us $(3)$:
$$\int \left(4-x(16-x^2)^{1/2}\right)\;dx = \int 4 dx - \int \color{red}{\bf \left(16-x^2\right)}^{1/2}\,\color{blue}{\bf x\,dx}\tag{2}$$
$$= \int 4 \;dx - \left(\color{blue}{\bf -\frac 12}\right)\int \color{red}{\bf U}^{1/2} \,\color{blue}{\bf dU}\tag{3}$$
That is, we compute the integrals: $$\int 4 \,dx + \left(\frac 12\right)\int U^{1/2} \, dU\quad = \quad 4x + \left(\frac 12\cdot \frac 23 \right) U^{3/2} + C$$
$$ = 4x+ \left(\frac 13\right)\left(16-x^2\right)^{3/2} + C\tag{4}$$

$(4)$ When using U-substitution, don't forget to "back-substitute"!
A: Since you have $$U=16-x^2$$
and $$dU=-2xdx$$ (Notice your mistake)
Continue what you are doing.
$$\int4dx+\frac{1}{2}\int U^{1/2}dU$$
A: As you said, define $u:=16-x^2$. The only thing wrong is the derivative, $\frac{du}{dx} = -2x \ dx$ Substituting will cause the $x$ to fall out.
