Integration: 1/sqrt(9-4x^2)

I am doing some integration drills and I am stuck with this one:

$\displaystyle

\int \frac {1}{\sqrt{9-4x^2}} dx

$

the solution is as follows:

$\displaystyle u = \frac {2}{3}x $

$\displaystyle du = \frac {2}{3} dx $

Step 3: $\displaystyle \frac {1}{3} \int \frac {1}{\sqrt{1 - (\frac {2x}{3})^2}} dx $

$\displaystyle = \frac {1}{2} \int \frac {1}{\sqrt{1-u^2}}du $

$\displaystyle

= \frac {1}{2} sin^-1 (\frac {2}{3}x) + C

$

What I don't understand is why $\displaystyle \frac {2x}{3}$ was chosen as u? and how they got step number 3?

Any help is appreciated.