Integral from (-pi/4) to (3 pi/4) of |sinx|?! Struggling, please help!?

When -π/4 ≤ x ≤ 0 : sin x ≤ 0, and thus |sin x| = -sin x
When 0 ≤ x ≤ π : sin x ≥ 0, and thus |sin x| = sin x
When π ≤ x ≤ 3π/4 : sin x ≤ 0, and thus |sin x| = -sin x

∫ sin(x) dx = -cos(x) + C

∫(-π/4→3π/4) |sin(x)| dx
= -∫(-π/4→0) sin(x) dx + ∫(0→π) sin(x) dx - ∫(π→3π/4) sin(x) dx
= [cos 0 - cos(-π/4)] - [cos(π) - cos(0)] + [cos(3π/4) - cos(π)]
= {1 - [(√2)/2]} - [-1 - 1] + {-[(√2)/2] - (-1)}
= 1 - [(√2)/2] + 1 + 1 - [(√2)/2] + 1
= 4 + √2

[ - π / 4 ] = [ - π / 4 , 0 ] U [ 0 , 3π / 4 ]...| sin x | = - sin x on the 1st integral and | sin x | = sin x on the 2nd integral...do the computations,,,,,note : the answer will be the same as if you had used [ 0 , π ]