Krdytkyu

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 $dx=frac{3}{2}sec^2theta dtheta$

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.

Probably the textbook first defines $u=2x$ and then $u=3tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3over 2}tan theta$$which by substitution leads to $$int{x^3over (4x^2+9)sqrt{4x^2+9}}dx=int {{27over 8}tan^3thetaover 27(1+tan^2theta)sectheta}cdot {3over 2}(1+tan^2theta)dtheta\={3over 16}int{tan^3thetaover sec theta}dtheta$$

You could avoid the extra substitution altogether (and greatly simplify other trig substitution problems) if you construct a right triangle to inform your substitutions.

Imagine a right triangle whose opposite (to $theta$) leg has length $2x$, adjacent leg has length $3$, and hypotenuse has length $sqrt{4x^2 + 9}$. From this I can deduce that $x = frac{3}{2}tantheta$, and therefore $dx = frac{3}{2}sec^2theta dtheta$. I can also determine that $sqrt{4x^2 + 9} = 3sectheta$. Substituting all this in gives:
$$
intfrac{x^3}{(4x^2+9)^frac{3}{2}} = int frac{left(frac{3}{2}tanthetaright)^3}{left(3secthetaright)^3} cdot frac{3}{2}sec^2theta dtheta = frac{3}{16}intfrac{tan^3theta}{sectheta}dtheta.
$$

The answers provided so far already outline exactly how to approach your question with trigonometric substitutions. Here I will speak more broadly about the common trig substitutions. The general goal is to take some rational function of multiple terms and reduce it down to a single term. For instance, here you want to have $4x^2 + 9$ to be represented by a single term.

This motivation of using trigonometry to reduce multiple terms to a single (or overall a much reduced amount) is quite common within integration. Although there are many different identities used, the majority are centred around the principle identity:

begin{equation}
sin^2(x) + cos^2(x) = 1
end{equation}

From here three core forms can be derived. The first is just a simple rearrangement:

begin{equation}
sin^2(x) + cos^2(x) = 1 rightarrow cos^2(x) = 1 - sin^2(x)
end{equation}

The next two are formed by dividing the principle identity through by $cos^2(x)$
begin{equation}
frac{1}{cos^2(x)}left(sin^2(x) + cos^2(x) right) = frac{1}{cos^2(x)} rightarrow tan^2(x) + 1 = sec^2(x)
end{equation}

Combining the three we have:

1. $sec^2(x) = tan^2(x) + 1$


2. $tan^2(x) = sec^2(x) - 1$


3. $sin^2(x) = 1 - cos^2(x)$

So here we two three different situations in which multiple terms can be reduced into a single term. These are (and to match the ordering of the list above)

1. $a^2 + x^2$


2. $x^2 - a^2$


3. $a^2 - x^2$

For your question you are working with (1).

Edit - Sorry I should have said that not only are you looking to replace multiple terms with a single term but you are also often looking to get rid of a 'square root'. You will see in the list above that in all three cases the resultant term is a squared term and thus when taking the square root you get rid of it!