Some things to watch out for:
- As already mentioned, if you substitute $u = 4x$, then you get $u^2 = (4x)^2 = 16x^2$ instead of $u^2 = 4x^2$. If you use $u = 2x$, then $u^2 = (2x)^2 = 4x^2$.
An annoying but important part of practicing good mathematics is being precise.
- In your integrals, you forgot the $dx$ and $du$ several times, which also lead you to forget to express $du$ in terms of $d\theta$. Perhaps you have heard of Riemann integrals; these may help you understand why the "$dx$" is there.
- You forgot the fact that if $F'(x) = f(x)$, then $\int f(x) dx = F(x) + C$. After all, when taking derivatives, all constants disappear, so the derivatives of $F(x)$ and $F(x) + 5$ are exactly the same. Thus, when calculating indefinite integrals, you do not know the constant term of the resulting function. So we generally write $\int f(x) dx = F(x) + C$ for some unknown constant $C$.
Here's a step-by-step approach of how I would do it. First, substitute $u = 2x$, so that $u^2 = (2x)^2 = 4x^2$, and apply $du = 2dx$:
$\int \sqrt{1 - \color{red}{4x^2}} \color{blue}{dx} = \int \sqrt{1 - \color{red}{u^2}} \color{blue}{\frac{du}{2}} = \frac{1}{2} \int \sqrt{1 - u^2} du.$
Then substitute $u = \sin \theta$ and $du = \cos \theta d\theta$ to get:
$\frac{1}{2} \int \sqrt{1 - \color{red}{u^2}} \color{blue}{du} = \frac{1}{2} \int \sqrt{1 - \color{red}{\sin^2 \theta}} \color{blue}{\cos \theta d\theta} = \frac{1}{2} \int \sqrt{\cos^2 \theta} \cos \theta d\theta = \frac{1}{2} \int \cos^2 \theta d\theta.$
Next, apply the double angle formula for the cosine, $\cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta)$, to get
$\frac{1}{2} \int \color{red}{\cos^2 \theta} d\theta = \frac{1}{2} \int \color{red}{\frac{1 + \cos 2\theta}{2}} d\theta = \frac{1}{4} \int (1 + \cos 2\theta) d\theta.$
Now split in two integrals and calculate those separately:
$\frac{1}{4} \int (\color{red}{1} + \color{blue}{\cos 2\theta}) d\theta = \frac{1}{4} \left(\int \color{red}{1} d\theta + \int \color{blue}{\cos 2\theta} d\theta\right) = \frac{1}{4}\left(\color{red}{\theta + C_1} + \color{blue}{\frac{\sin 2\theta}{2} + C_2}\right) \\ = \frac{\theta}{4} + \frac{\sin 2\theta}{8} + C.$
Applying a double angle formula for the sine, $\sin 2\theta = 2 \sin \theta \cos \theta$, we get
$\frac{\theta}{4} + \frac{\color{red}{\sin 2\theta}}{8} + C = \frac{\theta}{4} + \frac{\color{red}{2 \sin \theta \cos \theta}}{8} + C = \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} + C.$
Substituting back $\theta = \arcsin u$, and using $\cos (\arcsin u) = \sqrt{1 - u^2}$, we get
$\frac{\color{red}{\theta}}{4} + \frac{\sin \color{blue}{\theta} \cos \color{green}{\theta}}{4} + C = \frac{\color{red}{\arcsin u}}{4} + \frac{\sin(\color{blue}{\arcsin u}) \cos(\color{green}{\arcsin u})}{4} + C = \frac{\arcsin u}{4} + \frac{u \sqrt{1 - u^2}}{4} + C.$
Finally substituting $u = 2x$, you get
$\frac{\arcsin \color{red}{u}}{4} + \frac{\color{blue}{u} \sqrt{1 - \color{green}{u}^2}}{4} + C = \frac{\arcsin \color{red}{2x}}{4} + \frac{\color{blue}{2x} \sqrt{1 - \color{green}{(2x)}^2}}{4} + C = \boxed{\displaystyle\frac{\arcsin 2x}{4} + \frac{x \sqrt{1 - 4x^2}}{2} + C}.$