Understanding this calculus simplification I'm having a lot of trouble computing answers in the "Arc Length and Surfaces of Revolution", and I found it probably has to do with me not understanding the following kinds of simplication:
$$S= 2\pi\int x^3\sqrt{1+(3x^2)^2 } \, dx$$
$$= \frac{2\pi}{36}\int (36x^3)(1+9x^4)^\frac{1}{2} dx$$
Integrating:
$$ = \left.\frac{\pi}{18}\left[\frac{(1+9x^4)^\frac{3}{2}}{\frac{3}{2}}\right] \right|_a^b$$
What I don't understand here is why we divided $2\pi$ by 36 and multiplied $x^3$ by the same value. And how did that completely remove the $x^3$ when we went to integrate?
$ \left[\dfrac{(1+9x^4)^\frac{3}{2}}{\frac{3}{2}}\right]$ looks like the integral of only $ (1+9x^4)^\frac{1}{2} $
What am I missing here? What happened to the x cubed?
 A: In my opinion, the easiest way to understand this is through the idea of wishful thinking. That is a technique where you ask yourself "what would make this simple to do" and try to go that way.
In this case, you have the expression $(x^3)(1+9x^4)^\frac{1}{2} $ inside the integral. Now, if you know chain rule well enough, you're gonna think, ah, if I differentiate that bracket, that $9x^4$ is gonna get outside the bracket and become $36x^3$. We already have an $x^3$ there! So we surely can make this into a simple differentiation of the bracket.
So let's use that. We will rewrie it as $(36x^3)(1+9x^4)^\frac{1}{2} $ and just divide by 36 elsewhere. Now we know that this is pretty much what the differentiation of that bracket would be (again, thanks to knowing chain rule), except we're integrating and thus we need to raise the power by one and divide by that exponent, yielding the result.
Of course, substution works just as well, but that is not the way of thinking the example seems to suggest, and in many cases, it's an unnecessary extra step.
A: What you're missing is just the chain rule.
Integration by substitution is the chain rule in reverse.
The chain rule is differentiation by substitution.
$$
2\pi\int \sqrt{1+(3x^2)^2 } \,\underbrace{\Big(  x^3\,dx \Big)}_{\text{HINT}}
$$
A: This is just the Chain Rule. I guess it'd be easier for you to see why 36 appears if you substitute $u=1+9x^4$ so that $\mathrm{d}u = 36x^3\,\mathrm{d}x$. You can then rewrite the integral as
$$S = 2\pi \int x^3(1+9x^4)^{1/2} \, \mathrm{d}x = 2\pi \int \frac{1}{36}\sqrt{u} \, \mathrm{d}u.$$
A: Like the other answers have said, we need to use the chain rule (substitution) with $u=1+9x^4$ so that $du=36x^3\,dx$. As you can see, we need to get a $36$ in there! The way we do this is to multiply by $1=36/36$. Multiplying by $1$ doesn't change anything, but we can choose the denominator and numerator appropriately so that we get the missing number that we need.
