# Don't understand how to use trig sub on $\int\frac{x^3}{(4x^2+9)^\frac{3}{2}}dx$

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.

• \theta gives $\theta$ Jan 5 '19 at 21:26
• What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$? 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 $$\int{x^3\over (4x^2+9)\sqrt{4x^2+9}}dx=\int {{27\over 8}\tan^3\theta\over 27(1+\tan^2\theta)\sec\theta}\cdot {3\over 2}(1+\tan^2\theta)d\theta\\={3\over 16}\int{\tan^3\theta\over \sec \theta}d\theta$$

• 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. Jan 5 '19 at 23:35
• You're welcome. That's right. For example a $\sqrt{x^2-a^2}$ term requires a $x=a\cosh u$ substitution.... 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:

$$$$\sin^2(x) + \cos^2(x) = 1$$$$

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

$$$$\sin^2(x) + \cos^2(x) = 1 \rightarrow \cos^2(x) = 1 - \sin^2(x)$$$$

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

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!