Integrate by Trig Substitution $$\int_{3/4}^{3\sqrt{3}/4}\frac{x^3}{\sqrt{9-4x^2}}dx$$
So far I have the following:
$$ u=2x\Rightarrow$$
$$ u=a\sin \theta\Rightarrow 3\sin\theta$$
$$2x=3\sin\theta \rightarrow x=\frac32\sin\theta \rightarrow dx=\frac32\cos\theta d\theta$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\sqrt{9-9\sin^2\theta}}\frac32\cos\theta d\theta \Rightarrow$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\sqrt{9\cos^2\theta}}\frac32\cos\theta d\theta \Rightarrow$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{u^3}{2^32} d\theta \Rightarrow$$
$$\frac{1}{16}\int_{3/4}^{3\sqrt{3}/4}u^3 d\theta \Rightarrow$$
$$\frac{1}{16}\int_{3/4}^{3\sqrt{3}/4}27\sin^3\theta d\theta \Rightarrow$$
$$\frac{27}{16}\int_{3/4}^{3\sqrt{3}/4}\sin^3\theta d\theta \Rightarrow$$
$$$$
$$\int \sin^3\theta d\theta \Rightarrow \int (1-\cos^2\theta)\sin \theta d\theta$$
$$u = \cos \theta \rightarrow du=-\sin \theta d\theta \Rightarrow$$
$$\int (u^2-1)du \Rightarrow \frac{u^3}{3}-u \Rightarrow \frac{\cos^3\theta}{3}-\cos\theta $$
$$$$
$$\frac{27}{16}\bigg[\frac{\cos^3\theta}{3}-\cos\theta \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$
$$\frac{27}{16}\bigg[\frac{\cos^2\theta\cos\theta}{3}-\cos\theta \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$
$$\frac{27}{16}\bigg[\frac{1-4x^2\sqrt{1-\frac{4x^2}{9}}}{27}-\sqrt{1-\frac{4x^2}{9}} \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$
Am I doing this correctly?

You're complicating things with the double substitution. Consider
$$
\sqrt{9-4x^2}=3\sqrt{1-\frac{4x^2}{9}}
$$
and directly set $\frac{2}{3}x=\sin\theta$, or $x=\frac{3}{2}\sin\theta$, with $dx=\frac{3}{2}\cos\theta\,d\theta$.
If $x=\frac{3}{4}$, then you have $\frac{3}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{1}{2}$, while for $x=\frac{3\sqrt{3}}{4}$ you have $\frac{3\sqrt{3}}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{\sqrt{3}}{2}$.
Thus the integral becomes
$$
\int_{\pi/6}^{\pi/3}\frac{(27\sin^3\theta)/8}{3\cos\theta}\frac{3}{2}\cos\theta\,d\theta=
\frac{27}{16}\int_{\pi/6}^{\pi/3}\sin^3\theta\,d\theta
$$
This is computed by observing that
$$
\sin^3\theta=(1-\cos^2\theta)\sin\theta
$$
so you can set $\cos\theta=u$, so $-\sin\theta\,d\theta=du$ and the integral is
$$
\int_{\sqrt{3}/2}^{1/2}(u^2-1)\,du
$$
Your procedure is correct, but you carry around too much and doing small errors becomes easy.
