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 $\sin(2\theta)=2\sin(\theta)\cos(\theta).$

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 : $$=\frac{i}{8x^2} + c$$
