Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution! 
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$

So, make the substitution 
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x =  \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$
Am I making the correct substitutions here? Substituting $d\theta$ a quantity of $(\sqrt{1}\sec\theta)$ will cancel from the denominator. Somewhere along the line I need to use the identity $\sin(2\theta)=2\sin(\theta)\cos(\theta).$
 A: With the substitution $x=\frac {\sec \theta }{2}$ you get $dx = \frac {\sec \theta \tan \theta }{2} d\theta  $ and the integral changes to $$ \int \frac {4\tan^2 \theta \sec \theta }{ \sec ^3 \theta } d\theta =4  \int \frac {\tan^2 \theta  }{ \sec ^2 \theta } d\theta =4 \int \sin ^2 \theta d\theta$$
Now you can use the double angle equality which you mentioned.
A: Take $2x = \cos(\theta)$, then you will get 
$$\int\frac{\sqrt{\cos^2(\theta)-1}}{\cos^3(\theta)} d\theta= \int i\frac{\sin(\theta)}{\cos^3(\theta)} d\theta$$
then take another substitution. $t=\cos(\theta)$ to finally get : $$=\frac{i}{8x^2} + c$$
A: Hint:
If trigonometric substitution is not mandatory, for (where $a$ is an arbitrary constant )
$$\dfrac{\sqrt{4x^2+a}}{x^{2n-1}}=4^{n-1}\dfrac{\sqrt{4x^2+a}}{(4x^2)^n}\cdot4x$$
set $\sqrt{4x^2+a}=y,4x^2+a=y^2,y\ dy=4x\ dx$
A: Setting
$x=\frac{\sqrt{\theta^{2}+1}}{2}$
$dx=\frac{\theta}{2\sqrt{\theta^{2}+1}}$,
and the funcion to integrate is:
$\frac{4\theta^{2}}{(\theta^{2}+1)^{2}}$,
whose integral is:
$2. atan(\theta)-\frac{2\theta}{\theta^{2}+1}$.
Substituting the value of
$\theta=\sqrt{4x^{2}-1}$,
we get:
$I=2. atan(\sqrt{4x^{2}-1})-\frac{\sqrt{4x^{2}-1}}{2x^{2}}$.
