∫[sin(x/2)]^8dx=∫[35/128-cosx/4+7cos(2x)/32+cos(4x)/128-cosx(1-(sinx)^2)/4]dx(应用倍角公式)=∫[35/128-cosx/4+7cos(2x)/32+cos(4x)/128]dx-∫[cosx(1-(sinx)^2)/4]dx=35x/128-sinx/4+7sin(2x)/64+sin(4x)/51...