Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as  $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?

You can easily integrate with suitable substitution as follows
Let $2x=3\tan\theta\implies dx=\frac{3}{2}\sec^2\theta\ d\theta$
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac{(\frac32\tan\theta)^3}{(9\tan^2\theta+9)^{3/2}}\ \frac{3}{2}\sec^2\theta\ d\theta$$
$$=\left(\frac{3}{2}\right)^4\frac{1}{3^3}\int \frac{\tan^3\theta}{\sec^3\theta}\sec^2\theta\ d\theta$$
$$=\frac{3}{16}\int \frac{\tan\theta(\sec^2\theta-1)}{\sec\theta}\ d\theta$$
$$=\frac{3}{16}\int (\sec\theta\tan\theta-\sin\theta)\ d\theta$$
$$=\frac{3}{16}(\sec\theta+\cos\theta)+C$$
substitute back to $x$,
$$=\frac{3}{16}\left(\frac{\sqrt{4x^2+9}}{3}+\frac{3}{\sqrt{4x^2+9}}\right)+C$$
$$=\color{blue}{\frac{2x^2+9}{8\sqrt{4x^2+9}}}+C$$
