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.

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    $\begingroup$ \theta gives $\theta$ $\endgroup$ Jan 5 '19 at 21:26
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    $\begingroup$ What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$? $\endgroup$
    – Ennar
    Jan 5 '19 at 22:00

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 $$\int{x^3\over (4x^2+9)\sqrt{4x^2+9}}dx=\int {{27\over 8}\tan^3\theta\over 27(1+\tan^2\theta)\sec\theta}\cdot {3\over 2}(1+\tan^2\theta)d\theta\\={3\over 16}\int{\tan^3\theta\over \sec \theta}d\theta$$

  • $\begingroup$ So cases where the radicand's constants are not in the $\sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks. $\endgroup$
    – JSmith
    Jan 5 '19 at 23:35
  • $\begingroup$ You're welcome. That's right. For example a $\sqrt{x^2-a^2}$ term requires a $x=a\cosh u$ substitution.... $\endgroup$ Jan 6 '19 at 7:03

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 $x = \frac{3}{2}\tan\theta$, and therefore $dx = \frac{3}{2}\sec^2\theta d\theta$. I can also determine that $\sqrt{4x^2 + 9} = 3\sec\theta$. Substituting all this in gives: $$ \int\frac{x^3}{(4x^2+9)^\frac{3}{2}} = \int \frac{\left(\frac{3}{2}\tan\theta\right)^3}{\left(3\sec\theta\right)^3} \cdot \frac{3}{2}\sec^2\theta d\theta = \frac{3}{16}\int\frac{\tan^3\theta}{\sec\theta}d\theta. $$


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!


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