Explanation

(b)
Let $I = \int_{0}^{2n\pi}{\left\{ |\sin{x| - \frac{1}{2}|\sin{x|}} \right\} dx}$
$= \int_{0}^{2n\pi}\frac{1}{2}|\sin{x|dx}$
$= \frac{1}{2}\left\lbrack \int_{0}^{2\pi}{|\sin{x|dx + \int_{2\pi}^{4\pi}{|\sin{x|\ dx + ...}}}} \right.\$
$+ \int_{2(n - 1)\pi}^{2n\pi}{|\sin{x\} dx\left. \ \ \right\rbrack}}$
Now, $I_{1} = \int_{0}^{2\pi}{|\sin{x|dx}}$
$= \int_{0}^{\pi}{\sin{x\ dx - \int_{\pi}^{2\pi}{\sin{x\ dx}}}}$
$= \left\lbrack - \cos x \right\rbrack_{0}^{\pi} + \lbrack\cos{x\left. \ \ \right\rbrack_{\pi}^{2\pi}} = 4$
$\therefore\ I = \frac{1}{2}\lbrack 4 + 4 + 4 + ... + n\ times\rbrack$
$= \frac{1}{2}(4n) = 2n$
$I = \int_{0}^{2n\pi}{\frac{1}{2}|\sin{x|\ dx}}$
Since $|\sin\ x|$ is periodic with period $\pi.$
$\therefore\ I = 2n\int_{0}^{\pi}\frac{|\sin{x|dx}}{2}$
$I = n\int_{0}^{\pi}{\sin{x\ dx}}$
$= \lbrack - \cos{x\left. \ \ \right\rbrack_{0}^{\pi}}$
$\Rightarrow I = 2n$
Alternate
$I = \int_{0}^{2n\pi}{\frac{1}{2}|\sin{x|dx}}$
Since, $|\sin\ x|$is periodic with period $\pi.$
$\therefore I + 2n\int_{0}^{\pi}\frac{|\sin{x|}}{2}dx = n\int_{0}^{\pi}{\sin{x\ dx = n\lbrack - \cos{x\left. \ \ \right\rbrack_{0}^{\pi}}}}$
$\Rightarrow \ \ I = 2n$