Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$ 
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \theta = \frac{\sin \theta}{4}$$
Replace $\theta$ with $u$
$$\frac{\sin (\arcsin u)}{4} = \frac{u}{4} = \frac{4x}{4} = x$$
This is wrong and I have no idea why.

Let $2x = u$. So, $dx = \frac{1}{2}du$
$$\frac{1}{2}\int \sqrt{1-u^2} du$$
Replace $u = \sin \theta$, $du = cos \theta\ d\theta$
So, equation will be
$$\frac{1}{2}\int \sqrt{1-sin^2 \theta}\ cos\theta\ d\theta$$
$$\frac{1}{2}\int \sqrt{\cos^2\theta}\ cos\theta\ d\theta$$
$$\frac{1}{2}\int  cos^2\theta\ d\theta$$
Add and subtract $\frac{1}{4}\int d\theta$
$$\frac{1}{4}\int  2cos^2\theta\ d\theta - \frac{1}{4}\int d\theta + \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int  (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int  (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int  \cos2\theta\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{8}  (\sin2\theta) + \frac{1}{4} \theta + C$$
$$\frac{1}{8} (2\sin\theta.\cos\theta) + \frac{1}{4} \theta + C$$
As u = $\sin\theta$
$$\frac{1}{4} (u\sqrt{1-u^2}) + \frac{1}{4} \sin^{-1}u + C$$
As $2x = u$
$$\frac{1}{4} (2x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$
$$\frac{1}{2} (x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$
