U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$.  How did they get $u = 2x$ and not $u = x^3$?

Alternatively, change:
$$4x^2+9=u \Rightarrow x^2=\frac{u-9}{4}; xdx=\frac{du}{8};$$
$$\int_{9}^{36} \frac{\frac{u-9}{4}\cdot \frac{du}{8}}{u^{3/2}}=\frac{1}{32}\int_{9}^{36} \frac{1}{u^{1/2}}du-\frac{9}{32}\int_{9}^{36} \frac{1}{u^{3/2}}du=\cdots$$
