The correct option is
B π−2∫π0(1−|sin8x|)dx=∫π01.dx∫π/80sin8xdx+∫π/4π/8sin8xdx−∫3π/8π/4sin8xdx+∫π/23π/8sin8xdx−∫5π/8π/2sin8xdx+∫6π/85π/8sin8xdx−∫7π/86π/8sin8xdx+∫8π/87π/8sin8xdx
=[x]π0−[cos8x8]π/80+[−cos8x8]π/4π/8−[−cos8x8]3π/8π/4+[−cos8x8]π/23π/8−[−cos8x8]5π/8π/2+[−cos8x8]3π/45π/8−[−cos8x8]π/83π/4+[−cos8x8]π7π/8
=π−28−28−28−28−28−28−28−28
=π−2.
Hence, the answer is π−2.