Given f(x) = |sinx|

f(x) = |sinx| = sinx

f(-x) = |sin(-x)| = sinx

Hence f(x) = f(-x)

$\therefore$ |sinx| is even function

The Fourier series of an even function contains only cosine terms and is known as Fourier Series and is given by

$f(x) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^{\infty} a_{n}cosnx$

$a_{0} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)dx $ $a_{n}= \frac{2}{\pi}\int\limits_{0}^{\pi}f(x)cosnxdx$

$\therefore$ Let us first find

$a_{0} = \frac{2}{\pi}\int\limits_{0}^{\pi}sinxdx$

$\therefore a_{0} = \frac{2}{\pi}\int\limits_{0}^{\pi}sinxdx$

$\therefore a_{0} = \frac{2}{\pi}\big[-cosx\big]_{0}^{\pi} = \frac{2\c.2}{\pi} = \frac{4}{\pi}$

$\therefore a_{0} = \frac{4}{\pi}$

Now

$a_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}f(x)cosnxdx$

$\therefore a_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}sinx cosnx dx$

$\therefore a_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}sin(1 + n)x + sin(1 - n)x dx$

$\because sinA sinB = \frac{1}{2}[sin(A + B) + sin(A - B)]$

$\therefore a_{n} = \frac{1}{\pi}\int\limits_{0}^{\pi}sin(1 + n)x + sin(1 - n)xdx$

$\therefore a_{n} = \frac{1}{\pi}\bigg[\frac{-cos(1 + n)x}{1 + n} - \frac{cos(1 - n)x}{1 - n}\bigg]_{0}^{\pi}$

$\therefore a_{n} = \frac{1}{\pi}\bigg[\frac{-cos(1 + n)\pi}{1 + n} - \frac{cos(1 - n)\pi}{1 - n} + \frac{1}{1 - n} + \frac{1}{1 + n}\bigg]$

$\therefore a_{n} = \frac{1}{\pi}\bigg[\frac{-cos(\pi + n\pi)}{1 + n} - \frac{cos(\pi - n\pi)}{1 - n} + \frac{2}{1 - n^2}\bigg]$

$\therefore a_{n} = \frac{1}{\pi}\bigg[\frac{-(-cosn\pi)}{1 + n} - \frac{(-cosn\pi)}{1 - n} + \frac{2}{1 - n^2}\bigg]$

$\therefore a_{n} = \frac{1}{\pi}\bigg[\frac{2cosn\pi}{1 - n^2} + \frac{2}{1 - n^2}\bigg]$

$a_{n} = \frac{2}{\pi(1 - n^2)}(cosn\pi + 1)$

$a_{n} = \frac{2}{\pi(1 - n^2)}(2) $

If n is even

$a_{n} = \frac{4}{\pi(1 - n^2)}$

$a_{n} = 0$

If n is odd

$\therefore f(x) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^{\infty}a_{n}cosnx$

$\therefore f(x) = \frac{1}{2}*\frac{4}{\pi} + \sum\limits_{n = 1}^{\infty}a_{n}cos2nx$

($\because a_{n}$ exist only if n is even i.e. put n = 2n)

$\therefore f(x) = \frac{2}{\pi} + \sum\limits_{n = 1}^{\infty} \frac{4}{\pi(1 - (2n)^2)}cos2nx$

(i.e. put n = 2n)

$\therefore f(x) = \frac{2}{\pi}\frac{4}{\pi}\sum\limits_{n = 1}^{\infty}\frac{cos2nx}{(1 - 4n^2)}$

By re-arranging,

$f(x) = \frac{2}{\pi} - \frac{4}{\pi}\sum\limits_{n = 1}^{\infty}\frac{cos2nx}{(4n^2 - 1)}$