In my textbook it says
First we note that $(4x^2+9)^\frac{3}{2} = (\sqrt{4x^2+9})^3$ so trigonometric substitution is appropriate. Although $\sqrt{4x^2+9}$ is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution $u=2x$. When we combine this with the tangent substitution, we have $x=\frac{3}{2}\tan(\theta)$, which gives $dx=\frac{3}{2}\sec^2\theta d\theta$
I understand the basic trig substitutions in integrals, and I recognize that this is going to turn into some kind of $\tan$ substitution since the denominator $\sqrt{4x^2+9}$ is equivalent to $\sqrt{x^2+a^2}$, but for the life of me I cannot figure out why he is subbing $u=2x$. Does it have something to do with the fact that $\sqrt{4x^2}=2x$? After that, then he subs in $x=\frac{3}{2}\tan(\theta)$ which I don't see relating to anything except for that maybe when the $4$ gets factored out of the radical the $9$ somehow becomes $9/2$? The only relation to $\frac{3}{2}$ in this integral I see is the denominator's exponent, which I doubt is actually the reason for subbing $x=\frac{3}{2}\tan(\theta)$
I'm not asking for the entire problem to be solved, I know I can do that looking at my textbook as the parts after this "preliminary" step make sense, I just have no idea how to start this. If I'm missing some obvious algebraic step I'm sorry, but if anyone can shed some light on what's happening with the $u=2x$ and the $\tan$ sub, I would appreciate it so so much.
Best Answer
Probably the textbook first defines $u=2x$ and then $u=3\tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3\over 2}\tan \theta$$which by substitution leads to $$\int{x^3\over (4x^2+9)\sqrt{4x^2+9}}dx=\int {{27\over 8}\tan^3\theta\over 27(1+\tan^2\theta)\sec\theta}\cdot {3\over 2}(1+\tan^2\theta)d\theta\\={3\over 16}\int{\tan^3\theta\over \sec \theta}d\theta$$