I'm stuck in this one of trig substitution for fuctions. I got this:
$$\int\frac{dx}{\sqrt{(4x^2-9)^3}}.$$
I know that the answer is:
$$\frac{x}{9*\sqrt{4x^2-9}}+c.$$
And with the steps that I know about this type of substitution, I came up here, but.. I don´t know how to continue to the answer:
$$\frac{3}{2}\int \frac{\tan\theta \sec\theta}{(9\tan^2\theta)(3\tan\theta)} \,d\theta.$$
 A: Let 
$$I = \int\frac{dx}{\sqrt{(4x^2-9)^3}}$$
and make the substitution $x = \frac{3}{2} \, \operatorname{sec}\theta$ to obtain
\begin{align}
I &= \frac{3}{2} \, \int \frac{\operatorname{sec}\theta \, \tan\theta \, d\theta}{ 27 \, \tan^{3}\theta} \\
&= \frac{1}{18} \, \int \frac{\operatorname{sec}\theta}{\tan^{2}\theta} \, d\theta = \frac{1}{18} \, \int \frac{\cos\theta}{\sin^{2}\theta} \, d\theta \\
&= - \frac{1}{18} \, \frac{1}{\sin\theta} + c_{0}
\end{align}
Now, 
\begin{align}
\frac{2 \, x}{3} &= \frac{1}{\cos\theta} \to \cos\theta = \frac{3}{2 \, x} \\
\theta &= \cos^{-1} \left(\frac{3}{2 \, x}\right) \\
\sin\theta &= \frac{1}{\sqrt{1 - \left(\frac{3}{2 \, x}\right)^{2}}} = \frac{2 \, x}{\sqrt{4 \, x^2 - 9}}
\end{align}
leading to 
$$I = - \frac{1}{9} \, \frac{x}{\sqrt{4 \, x^2 - 9}} + c_{0}.$$
A: Hint: Expand the $\tan\theta$ and $\sec\theta$ to their respective $\sin\theta$, $\cos\theta$ parts and cancel. You are left with an integral that can be solved using a $u$-substitution.
A: HINT...hyperbolic functions make this job easier. try substituting $2x=3\cosh\theta$
A: Even though hyperbolic method is a good one, but I think there is one even simpler method,
just take x^2 out of the radical then substitute
(4-9/(x^2))=(t^2).
