Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses? 


When you're doing the trigonometric substitution, you write $x=a\sin\theta$, which is good; you should also remember how to get back from $\theta$ to $x$, that is,
$$
\theta=\arcsin\frac{x}{a}=\arcsin\frac{x}{1/2}=\arcsin(2x)
$$
which actually should be the starting point, because it guarantees the angle $\theta$ is between $-\pi/2$ and $\pi/2$.
When you arrive to
$$
\frac{1}{4}\theta+\frac{1}{8}\sin(2\theta)=
\frac{1}{4}\theta+\frac{1}{4}\sin\theta\cos\theta
$$
you indeed need to get back to $x$. Since $-\pi/2\le\theta\le\pi/2$, you know $\cos\theta\ge0$ and so
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-4x^2}
$$
In conclusion your integral is
$$
\frac{1}{4}\arcsin(2x)+\frac{1}{2}x\sqrt{1-4x^2}
$$

However, there's no need for trigonometric substitutions. Consider
$$
\int\sqrt{1-t^2}\,dt=
\int\frac{1-t^2}{\sqrt{1-t^2}}\,dt=
\int\frac{1}{\sqrt{1-t^2}}\,dt+\int\frac{-t^2}{\sqrt{1-t^2}}\,dt
$$
The first one is immediate; the second one can be computed with integration by parts:
$$
\int t\frac{-t}{\sqrt{1-t^2}}\,dt=
t\sqrt{1-t^2}-\int\sqrt{1-t^2}\,dt
$$
All in all, we have
$$
\int\sqrt{1-t^2}\,dt=
\arcsin t+t\sqrt{1-t^2}-\int\sqrt{1-t^2}\,dt
$$
so we can transport the integral from the right-hand side to the left-hand side and get
$$
\int\sqrt{1-t^2}\,dt=
\frac{1}{2}\arcsin t+\frac{1}{2}t\sqrt{1-t^2}
$$
For your integral use the substitution $2x=t$.
