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

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.
