<s>[INST] 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? 

 [/INST] In general case whenever you have something like $1-a^2x^2$. Remember $\sin^2x+\cos^2x=1$ and so we should multiply it by $a$. Thus take $\sin y=\frac{1}{a}x$. Now we have
$\sqrt{1-a^2x^2}=\sqrt{1-\sin^2y}=\cos y$.
Hence one can see that 
$\int\sqrt{1-a^2x^2}dx=\int\sqrt{1-\sin^2y}\times a\cos y \,dy=\int a\cos^2 y dy$
Now we have $\int \cos^2y$. The standard technique to deal with this is using the gold formula $\cos^2y=\frac{1}{2}(1+\cos 2y)$. In fact the gold formula is revealing the connection between $\cos^2$ and $\cos$.
The most controversial part is the integral of $\int\cos 2ydy$.
It is $\frac{1}{2}\sin 2y$ or in other words it is equal to $\sin y\cos y$. Now recall that $\sin y=\frac{1}{2}x$. Thus one can see that $\cos y=\sqrt{1-\frac{1}{4}x^2}$ and we are done, as we only need to have $\sin y$ and $\cos y$.
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