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? 


This is called Trigonometric Substitution.
$$\int\sin^2 x\cos xdx, \color{green}{\text{let }u=\sin x}\implies du=\cos xdx$$
$$\int\sin^2 x(\cos xdx)=\int u^2du =\frac{u^3}{3}+C =\frac{\sin^3 x}{3}+C$$
Whereas this is called Inverse Trigonometric Substitution
$$\int\sqrt{1-4x^2}dx = \int\sqrt{1-(2x)^2}dx, \color{green}{\text{let }\sin u=2x}\implies \cos u du=2dx$$
$$\begin{array}{lll}
\int\sqrt{1-(2x)^2}dx&=&\frac{1}{2}\int 2\sqrt{1-(2x)^2}dx\\
&=&\frac{1}{2}\int \cos u\sqrt{1-\sin^2 u}du\\
&=&\frac{1}{2}\int \cos^2 udu\\
&=&\dots\\
\end{array}$$
Notice that the substitution $\sin u = 2x$ can be rewritten as
$$u = \sin^{-1} (2x)$$
Compare the differences between the substitutions in the two examples, and it should become clear why inverses are necessary.
It may be instructive to use implicit differentiation to derive
$$\frac{d}{dt}\sin^{-1}t = \frac{1}{\sqrt{1-t^2}}$$
and then explicitly make the substitution $u = \sin^{-1}(2x)$
$$\frac{du}{dx} = \frac{d}{d2x}\sin^{-1}(2x)\cdot\frac{d}{dx}2x = \frac{2}{\sqrt{1-(2x)^2}}$$
$$du =  \frac{2dx}{\sqrt{1-(2x)^2}}$$
Getting back to our integral
$$\int\sqrt{1-(2x)^2}dx=\int\sqrt{1-(2x)^2}\cdot\frac{\color{green}{2}\sqrt{1-(2x)^2}}{2\color{green}{\sqrt{1-(2x)^2}}}dx = \int\frac{1}{2}(1-(2x)^2)\cdot\color{green}{\frac{2}{\sqrt{1-(2x)^2}}}dx$$
$$= \int\frac{1}{2}(1-\sin^2(\sin^{-1}(2x)))\cdot\frac{2}{\sqrt{1-(2x)^2}}dx$$
$$= \int\frac{1}{2}(1-\sin^2 u)du = \frac{1}{2}\int\cos^2 udu=\dots$$
