Question

 integral of modulous of cosx in interval (0,2pi)

Answer 1

since cos x must always be positive
it just area under cosine function
so integral
$4\int\limits^\frac{ \pi }{2}_ 0 {cos (x)} \, dx=4$

Answer 2

The graph of y = |cos x| over the interval [0, 2pi] is exactly two "humps" of the cosine curve (see picture). So, we just need to integrate over one hump, then double the result: 
INT |cos x| dx (from 0 to 2pi) 
= 2 * INT cos x dx (from -pi/2 to pi/2, because this would be an interval for one hump) 
= 2 * sin x, evaluated from -pi/2 to pi/2 
= 2 * [sin(pi/2) - sin(-pi/2)] 
= 2 * [1 - (-1)] 
= 2 * 2 
= 4