<s>[INST] 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 $u = x^3$?
 [/INST] When $u = 2x$, $du = 2dx$,
$$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx
= \int^{3\sqrt{3}}_0\frac{u^3/8}{(u^2+9)^{3/2}}\frac{du}{2},
$$
this allows us to do further substitution $u = 3 \tan t$ to get rid of the root sign in the denominator.
If you use $u = x^3$, $du = 3x^2 dx$,
$$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx
= \int^{81\sqrt{3}/8}_0\frac{\sqrt[3]{u}/3}{(u^{2/3}+9)^{3/2}}du,
$$
which is more complicated then the first equality as this involves the cubic root.
</s>