<s>[INST] 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.
 [/INST] $$\int \sqrt{1-4x^2}\ dx$$
Substitute
$$\begin{align*}x &= \frac{\sin (\theta)}{2}\\ dx &= \frac{\cos(\theta)}{2}d \theta\end{align*}$$
So the intregral become
$$\begin{align*}
\int \sqrt{1-4x^2}\ dx &=\frac{1}{2}\int \cos^2(\theta)d\theta\\
&=\frac{1}{2}\int \frac{\cos(2\theta) + 1}{2}d\theta\\
&=\frac{\sin(2\theta)}{8}+\frac{\theta}{4}
\end{align*}$$
Again 
$$\theta = \arcsin(2x)$$
And
$$\begin{align*}
\sin 2\theta &= 2\sin \theta \cos \theta\\
             &=4x\sqrt{1-4x^2}
\end{align*}$$
So the answer is
$$\frac{x\sqrt{1-4x^2}}{2}+\frac{\arcsin(2x)}{4}+C$$
</s>