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? 

 A: 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 $\cos y$.
A: 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 $2x=t$.
A: $$\int\sqrt{1-4x^2}\space\space\text{d}x=$$

Substitute $x=\frac{\sin(u)}{2}$ and $\text{d}x=\frac{\cos(u)}{2}\space\space\text{d}u$. Then $\sqrt{1-4x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\sin^{-1}(2x)$:

$$\frac{1}{2}\int\cos^2(u)\space\space\text{d}u=$$
$$\frac{1}{2}\int\left(\frac{1}{2}\cos(2u)+\frac{1}{2}\right)\space\space\text{d}u=$$
$$\frac{1}{4}\int\cos(2u)\space\space\text{d}u+\frac{1}{4}\int 1\space\space\text{d}u=$$

Substitute $s=2u$ and $\text{d}s=2\space\space\text{d}u$:

$$\frac{1}{8}\int\cos(s)\space\space\text{d}s+\frac{1}{4}\int 1\space\space\text{d}u=$$
$$\frac{\sin\left(s\right)}{8}+\frac{u}{4}+\text{C}=$$
$$\frac{\sin\left(2u\right)}{8}+\frac{u}{4}+\text{C}=$$
$$\frac{\sin\left(2\sin^{-1}(2x)\right)}{8}+\frac{\sin^{-1}(2x)}{4}+\text{C}$$
A: 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
$$\int\sqrt{1-(2x)^2}dx=\int\sqrt{1-(2x)^2}\cdot\frac{\color{green}{2}\sqrt{1-(2x)^2}}{2\color{green}{\sqrt{1-(2x)^2}}}dx = \int\frac{1}{2}(1-(2x)^2)\cdot\color{green}{\frac{2}{\sqrt{1-(2x)^2}}}dx$$
$$= \int\frac{1}{2}(1-\sin^2(\sin^{-1}(2x)))\cdot\frac{2}{\sqrt{1-(2x)^2}}dx$$
$$= \int\frac{1}{2}(1-\sin^2 u)du = \frac{1}{2}\int\cos^2 udu=\dots$$
