Last year, I was trying to solve a trigonometric integral with absolute values depending on any positive integer $n$. After trying multiple methods, I decided to go with a slightly easier integral to see if I could gain some useful tools. The following is an attempt at proving a proposal I made up.
Proposal. Let $n$ be a positive integer greater than $1$. Prove
$$\int_{0}^{2\pi}\left|\sin{(n-1)x}\right|dx = 4.$$
Proof. Let $f(x) = \sin{(n-1)x}$ and $I$ equal the given integral. The roots of $f(x)$ are obtained by
$$\sin{((n-1)x)} = 0 \implies (n-1)x = \pi k \implies x = \frac{\pi k}{n-1},$$
where $k$ is in some subset of $\mathbb{Z}$. To find that set, we can bound $k$ since $x \in \left[0,2\pi\right]$ like this:
$$0 \leq \frac{\pi k}{n-1} \leq 2\pi \iff 0 \leq k \leq 2n-2.$$
The set of roots of $f(x)$ is $\left\{\frac{\pi k}{n-1} \in \mathbb{R} : n \in \mathbb{N}\setminus\left\{1\right\} \text{ and } k \in \left\{0,1,2,\ldots,2n-2\right\}\right\}$. We will partition $I$ such that the lower and upper bounds are elements of the aforementioned set. We will make $I$ equal to two summations adding each other where the first one represents the sum of positive areas and the second one represents the sum of the other areas (that would have been negative if the absolute values weren't involved). To elaborate,
$$I = \sum_{i=0}^{n-2}\int_{\frac{2\pi i}{n-1}}^{\frac{\left(2i+1\right)\pi}{n-1}}f\left(x\right)dx\ +\ \sum_{i=0}^{n-2}\int_{\frac{\left(2i+1\right)\pi}{n-1}}^{\frac{2\pi i}{n-1}}-f\left(x\right)dx\ =\ 2\sum_{i=0}^{n-2}\int_{\frac{2\pi i}{n-1}}^{\frac{\left(2i+1\right)\pi}{n-1}}f\left(x\right)dx.$$
To further prove where this equation comes from, we will prove via induction that
$$\int_{0}^{2\pi}\left|\sin\left(\left(n-1\right)x\right)\right|dx\ =\ \ 2\sum_{i=0}^{n-2}\int_{\frac{2\pi i}{n-1}}^{\frac{\left(2i+1\right)\pi}{n-1}}\sin\left(\left(n-1\right)x\right)dx.$$
Base Step. If $n = 2$, then the left side becomes
$$\int_{0}^{2\pi}\left|\sin\left(\left(n-1\right)x\right)\right|dx = \int_{0}^{2\pi}\left|\sin\left(\left(2-1\right)x\right)\right|dx = 4.$$
The right side becomes
$$2\sum_{i=0}^{n-2}\int_{\frac{2\pi i}{n-1}}^{\frac{\left(2i+1\right)\pi}{n-1}}\sin\left(\left(n-1\right)x\right)dx = 2\sum_{i=0}^{2-2}\int_{\frac{2\pi i}{2-1}}^{\frac{\left(2i+1\right)\pi}{2-1}}\sin\left(\left(2-1\right)x\right)dx = 4.$$
Inductive Step. Fix $n = k$ and suppose
$$\int_{0}^{2\pi}\left|\sin\left(\left(k-1\right)x\right)\right|dx\ =\ \ 2\sum_{i=0}^{k-2}\int_{\frac{2\pi i}{k-1}}^{\frac{\left(2i+1\right)\pi}{k-1}}\sin\left(\left(k-1\right)x\right)dx.$$
We want to show
$$\int_{0}^{2\pi}\left|\sin\left(kx\right)\right|dx\ =\ \ 2\sum_{i=0}^{k-1}\int_{\frac{2\pi i}{k}}^{\frac{\left(2i+1\right)\pi}{k}}\sin\left(kx\right)dx.$$
We will manipulate the right side of that equation above like this:
$$\eqalign{ 2\sum_{i=0}^{k-1}\int_{\frac{2\pi i}{k}}^{\frac{\left(2i+1\right)\pi}{k}}\sin\left(kx\right)dx &= 2\left(\sum_{i=0}^{k-2}\int_{\frac{2\pi i}{k}}^{\frac{\left(2i+1\right)\pi}{k}}\sin\left(kx\right)dx\ +\ \int_{\frac{2\pi\left(k-1\right)}{k}}^{\frac{\left(2k-1\right)\pi}{k}}\sin\left(kx\right)dx\right) \cr &= \frac{2\left(k-1\right)}{k}\sum_{i=0}^{k-2}\int_{\frac{2\pi i}{k-1}}^{\frac{\left(2i+1\right)\pi}{k-1}}\sin\left(\left(k-1\right)u\right)du+\frac{4}{k} \text{ (after letting $kx = (k-1)u$)} \cr &= \frac{k-1}{k}\int_{0}^{2\pi}\left|\sin\left(\left(k-1\right)x\right)\right|dx+\frac{4}{k} \text{ (by inductive hypothesis)} \cr &= \frac{k-1}{k}\int_{0}^{\frac{2\pi\left(k-1\right)}{k}}\left|\sin\left(kv\right)\right|\left(\frac{k}{k-1}\right)dv+\frac{4}{k} \text{ (after letting $(k-1)x = kv$)} \cr &= \int_{0}^{2\pi-\frac{2\pi}{k}}\left|\sin\left(kv\right)\right|dv+\frac{4}{k} \cr &= \int_{0}^{2\pi}\left|\sin\left(kv\right)\right|dv\ -\ \int_{2\pi-\frac{2\pi}{k}}^{2\pi}\left|\sin\left(kv\right)\right|dv\ +\ \frac{4}{k} \cr &= \int_{0}^{2\pi}\left|\sin\left(kv\right)\right|dv+\int_{0}^{-\frac{2\pi}{k}}\left|\sin\left(kv\right)\right|dv+\frac{4}{k} \cr &= \int_{0}^{2\pi}\left|\sin\left(kv\right)\right|dv+\frac{1}{k}\int_{0}^{-2\pi}\left|\sin\left(w\right)\right|dw+\frac{4}{k} \text{ (after letting w = kv)} \cr &= \int_{0}^{2\pi}\left|\sin\left(kx\right)\right|dx. }$$
Therefore by Principal of Mathematical Induction, the induction proof is done.
Going back to the solution, we see
$$\eqalign{ I &= 2\sum_{i=0}^{n-2}\int_{\frac{2\pi i}{n-1}}^{\frac{\left(2i+1\right)\pi}{n-1}}\sin\left(\left(n-1\right)x\right)dx \cr &= 2\sum_{i=0}^{n-2}\frac{1}{n-1}\left(\cos\left(2\pi i\right)-\cos\left(\left(2i+1\right)\pi\right)\right) \cr &= 2\sum_{i=0}^{n-2}\frac{1}{n-1}\left(1+1\right) \cr &= \frac{4}{n-1}\sum_{i=0}^{n-2}1 \cr &= \frac{4}{n-1}\left(n-2+1\right) \cr &= 4. }$$
Q.E.D.
Is there a nicer way to prove that proposal? I know it's a lot of work but if anyone can find any optimizations/errors then please let me know. Or if you have any other ideas/solutions you want to add, I would greatly appreciate it!
A more general result: let $m$ and $n$ be positive integers. If $f$ is any periodic function of period $T$, by changing variable $t=nx$ we find \begin{align} \int_{0}^{mT} f(nx) dx&=\frac{1}{n}\int_{0}^{mnT} f(t)dt= \frac{1}{n}\sum_{k=1}^{mn}\int_{(k-1)T}^{kT} f(t)dt\\ &=\frac{1}{n}\sum_{k=1}^{mn}\int_{0}^{T} f(t)dt=\frac{mn}{n}\int_{0}^{T} f(t) dt=m\int_{0}^{T} f(t) dt \end{align} where we apply: Show that $\int_a^{a+T}f(t)dt=\int_0^Tf(t)dt$ for all $a$ when $f$ is $T-$periodic.
Notice that the final result does not depend on $n$.
When $f(x)=|\sin(x)|$ then $T=\pi$ and $$\int_{0}^{T} f(t) dt=\int_{0}^{\pi} |\sin(t)| dt=\int_{0}^{\pi} \sin(t) dt=[-\cos(t)]_0^\pi=2.$$ Therefore $$\int_{0}^{m\pi} |\sin(nx)| dx=2m.$$