what is the integral of $\displaystyle 1/\sqrt (36-16x^2)$
use substitution
$\displaystyle \displaystyle \frac{1}{\sqrt{36-16x^2}}=\frac{1}{4}\frac{1}{\sqrt{\left( \frac{6}{4}\right)^2-x^2}}= \frac{1}{4}\frac{1}{\sqrt{\left( \frac{3}{2}\right)^2-x^2}}$
So the question you need to ask yourself is what trig sub works for things of the form
$\displaystyle \displaystyle \frac{1}{\sqrt{a^2-x^2}}$