I was recently playing around with some easy volume by revolution problems that I just randomly make up for fun, and I found a weird and interesting pattern that I can't easily (or otherwise) explain. Say I want to find the volume of one cycle of $\sin{x}$, revolved about the x axis. I made up this problem because I assumed the net signed area to be 0, but I wanted to see if that was a misconception, and if so, correct it (it actually does turn out to be 0, as far as I can tell, so yay! edit: I calculated the wrong integral by mistake. It's not 0. oops). I came up with the following equation for volume: $$ V = \pi\int_0^{2\pi} \sin^2{x}\, \mathrm{d}x $$ Now, if I forgot how to integrate $\sin^2{x}$, I might try to use a trig identity and change the above to: $$ \pi\int_0^{2\pi} (1-\cos^2{x})\, \mathrm{d}x $$ Here's where I thought things got interesting. I can split up that integral, and write it as: $$ \pi\int_0^{2\pi} 1\,\mathrm{d}x - \pi\int_0^{2\pi} \cos^2{x}\, \mathrm{d}x $$ Now let's say I was really desperate, and tried to use an identity again by replacing $\cos^2{x}$ with $1-\sin^2{x}$. If I kept following the pattern, I'd get this (I think): $$ V = \pi\int_0^{2\pi} 1\,\mathrm{d}x - \pi\int_0^{2\pi} 1\,\mathrm{d}x + \pi\int_0^{2\pi} \sin^2{x}\, \mathrm{d}x $$ $$ V = \pi\int_0^{2\pi} 1\,\mathrm{d}x - \pi\int_0^{2\pi} 1\,\mathrm{d}x - \pi\int_0^{2\pi} 1\,\mathrm{d}x - \pi\int_0^{2\pi} \cos^2{x}\, \mathrm{d}x $$
And $\pi\int_0^{2\pi} 1\,\mathrm{d}x$ is definitely not 0 (it's actually $2\pi^2$ I think), or some other value which easily explains this pattern. Perhaps if we took the sum as n number of those terms approaches infinity? I would attempt this, but I've never taken the sum of integrals before, and I don't trust my work to be correct. In fact, I don't trust my work even to this point to be necessarily correct, so I was hoping that maybe some insights from someone else could shed some light on this. Thanks!
This is what it's supposed to be,
$$V=\pi\int\limits_0^{2\pi}1\,\mathrm{d}x-\pi\int\limits_0^{2\pi}1\,\mathrm{d}x\color{red}{+}\pi\int\limits_0^{2\pi}1\,\mathrm{d}x-\pi\int\limits_0^{2\pi}\cos^2x\, \mathrm{d}x$$
Now, to clear the confusion, what you're getting (if you continue that kinda decomposition) is,
$$\frac{V}{\pi}=\begin{cases}\displaystyle\left(\sum_{i=0}^n \int\limits_0^{2\pi}(-1)^i\,\mathrm{d}x\right)-\int\limits_0^{2\pi}\cos^2 x\,\mathrm{d}x~,~\textrm{if }n\textrm{ is even}\\ \displaystyle\left(\sum_{i=0}^n \int\limits_0^{2\pi}(-1)^i\,\mathrm{d}x\right)+\int\limits_0^{2\pi}\sin^2 x\,\mathrm{d}x~,~\textrm{if }n\textrm{ is odd}\end{cases}$$
By telescopy, it simply reduces to,
$$\frac{V}{\pi}=\begin{cases}\displaystyle \int_0^{2\pi}1\,\mathrm{d}x-\int\limits_0^{2\pi}\cos^2 x\,\mathrm{d}x~,~\textrm{if }n\textrm{ is even}\\ 0+\displaystyle\int\limits_0^{2\pi}\sin^2 x\,\mathrm{d}x~,~\textrm{if }n\textrm{ is odd}\end{cases}$$
Both of these result in the same thing using the identity $\cos^2 x + \sin^2 x = 1$ and that same thing is,
$$\frac{V}{\pi}=\int_{0}^{2\pi}\sin^2\,\mathrm{d}x\implies V=\pi\int_{0}^{2\pi}\sin^2 x\,\mathrm{d}x$$
which is the same stuff you started with. You were just reversely using the technique of telescopy to create a conditionally convergent sum. Of course, you don't get anything infinite out of it. The integral $V$ is finite and you can see the value here.