Can a limit be changed twice when evaluating an integral? I am asked to evaluated
$ \int^{\frac{3\sqrt3}{2}}_0 \frac{x^3}{(4x^2+9)^{\frac{3}{2}}} $ 
where $a>0$
the text book shows that during the u-substitution you should get
$\int^{3\sqrt3}_0 {\frac{{\frac{1}{2}}(u^3)}{\sqrt(u^2+9)^3}\frac{1}{2}}du $
then the limit is changed again after doing a trig substitution
What was done to change the limit at the step shown?
 A: I guess the suggested substitution was $u^2=4x^2$, so that $u=2x$. From thereone you got $u_b=2\cdot 0$ for the limit on the bottom and $u_t=2\cdot \frac{3\sqrt{3}}{2}=3\sqrt{3}$ for the top. With the change of the differential by $dx=\frac{du}2$ the integral becomes
$$\int_0^{\frac{3\sqrt{3}}{2}} \frac{x^3}{(4x^2+9)^{\frac32}} dx~=~ \int_0^{3\sqrt{3}} \frac{\left(\frac{u}2\right)^3}{(u^2+9)^{\frac32}} \frac{du}2$$
When I am not mistaken this integral is different to the one you offer by a factor of $\frac14$ and I am not sure why.

You can dodge the trigonometric substitution by setting $t=4x^2+9$ $(x^2=\frac14(t-9))$ and therefore $xdx=\frac{dt}8$, $t_b=9$, $t_t=36$ so that the integral changes to
$$\int_0^{\frac{3\sqrt{3}}{2}} \frac{x^3}{(4x^2+9)^{\frac32}} dx~=~\int_6^{36} \frac{\frac14(t-9)}{t^{\frac32}} \frac{dt}8~=~\frac1{32}\int_6^{36}\frac1{t^{\frac12}}-\frac9{t^{\frac32}}dt$$
where the last integral only needs you to integrate roots which is quite easy.
A: The first substition is $$u=2x\to du=2dx\to \frac12 du=dx$$
This leads to:
$$\int_{\alpha}^{\beta}{\frac{(\frac u2)^3}{(u^2+9)^3}\cdot\frac12 du}$$
(Note my use of $\alpha$ and $\beta$ in the limits - it is my strong advice that you deal with limits after substituting lest you confuse yourself - this is how I do it too!)
Your limits $dx$ are $0$ and $\frac32\sqrt3$ respectively. Since $u=2x$, your limits $du$ are $0$ and $3\sqrt3$ respectively.
When the trigonometric substitution is made (I don't know which would be used here - but for example's sake let's say it is $t=\sin(u)$), then you would apply that function to your limits to get your new limits (with my example your limits $dt$ would be $\sin (0)=0$ and $\sin (3\sqrt3)\approx-0.8852$ respectively)
