<s>[INST] 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?
 [/INST] The OP correctly identified the power rule for exponents
$$(b^n)^m= b^{n\times m}=b^{nm}.$$
Therefore for $b=\sec\theta,n=2,$ and $m=3/2$,
$$27\ ( \sec^2 \theta ) ^\frac{3}{2}=27(\sec\theta)^{2\times\frac32}=27(\sec^3\theta).$$
To finish the problem, the OP can apply the substitution $x = \frac{3}{2} \tan \theta \implies dx = \frac{3}{2} \sec^2 \theta \,d\theta.$
$$\int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx=\int\frac{\frac{27}{8}\tan^3\theta}{27\sec^3\theta}\frac{3}{2} \sec^2 \theta \,d\theta$$
$$=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\sec^2 \theta \,d\theta$$
$$=\frac{3}{16}\int\frac{\tan^3\theta}{\sec\theta} \,d\theta$$
$$=\frac{3}{16}\int\frac{(\sec^2\theta-1)\tan\theta}{\sec\theta} \,d\theta$$
$$=\frac{3}{16}\left(\int\,d(\sec \theta)+\int\,d(\cos\theta)\right)$$
$$=\frac{3}{16}\left(\sec \theta+\cos\theta\right)+C.$$
The substitution $x = \frac{3}{2} \tan \theta \implies \sqrt{4x^2+9}=3\sec\theta$ which forms
$$=\frac{3}{16}\left(\frac{\sqrt{4x^2+9}}{3}+\frac{3}{\sqrt{4x^2+9}}\right)+C$$
$$=\frac{3}{16}\left(\frac{4x^2+18}{3\sqrt{4x^2+9}}\right)+C$$
$$=\boxed{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$

As an alternative approach, I observed that
$$x^3=\frac{x(4x^2+9)-9x}{4},$$
therefore
$$\int \frac{x^3}{(4x^2+9)^{3/2}}\,dx=\frac{1}{4}\int \frac{x(4x^2+9)-9x}{(4x^2+9)^{3/2}}\,dx$$
$$=\frac{1}{4}\int \frac{x}{\sqrt{4x^2+9}}\,dx-\frac{9}{4}\int\frac{x}{(4x^2+9)^{3/2}}\,dx$$
Substituting $u=4x^2+9 \implies du=8x\,dx$ on both integrals forms
$$=\frac{1}{32}\int\frac{1}{\sqrt{u}}\,du-\frac{9}{32}\int\frac{1}{u^{3/2}}\,du$$
$$=\frac{1}{16}\sqrt{u}+\frac{9}{16\sqrt{u}}+C$$
$$=\frac{1}{16}\left(\frac{u+9}{\sqrt{u}}\right)+C$$
$$=\frac{1}{16}\left(\frac{4x^2+18}{\sqrt{4x^2+9}}\right)+C$$
$$=\boxed{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$
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