Today I came across the following problem:
Evaluate$$\displaystyle\int_0^{2\pi} [|\sin(x)| + |\cos(x)|]\ dx $$ where $[.]$ denotes greatest integer function.
I'm a little confused in simplifying the function $[|\sin (x)| + |\cos(x)|]$.
$\because -1\le \sin(x) \le 1$ and $-1 \le \cos(x) \le 1$
This implies that $ 0 \le|\sin x| \le 1$ and $0 \le|\cos x|\le 1$
Now adding both these equations will give, $$0 \le |\sin x| + |\cos x | \le 2$$
Max. & min. values of $\sin(x) + \cos(x)$ are $\sqrt{2}$ & $1$.So maybe $1 \le |\sin x| + |\cos x | \le \sqrt{2}$? $\tag{$*$}$
This will give $[|\sin x| + |\cos x|] = 1$.
$$\therefore \int_0^{2\pi}[|\sin x| + |\cos x|]\ dx = 2\pi. $$
I'm not sure what I've done in step $(*)$ is right or wrong.
$|\sin x|\geq \sin^2 x,$ and $|\cos x|\geq\cos^2 x.$
So $$|\sin x|+|\cos x|\geq \sin^2x+\cos^2x=1.$$
Also:
$$\begin{align}(|\sin x|+|\cos x|)^2&=\sin^2x+\cos^2x+|2\sin x\cos x|\\&=1+|\sin2x|\\&\leq 2.\end{align}$$
So your integrand is identically $1,$ and the integral is $2\pi.$