I'm stuck in this one of trig substitution for fuctions.

Question

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:

$$\frac{3}{2}\int \frac{\tan\theta \sec\theta}{(9\tan^2\theta)(3\tan\theta)} \,d\theta.$$

Answer 1

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 $$I = - \frac{1}{9} \, \frac{x}{\sqrt{4 \, x^2 - 9}} + c_{0}.$$

Answer 2

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.

Answer 3

HINT...hyperbolic functions make this job easier. try substituting $2x=3\cosh\theta$

Answer 4

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).