∫（sinx）^2dx,等于（1/2）x-（1/4）sin2x,到底是怎么得出来的?

三角函数公式：
sin2x = 2sinxcosx
cos2x = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x
sin²x = (1 - cos2x)/2,cos²x = (1 + cos2x)/2
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∫ sin²x dx
= (1/2)∫ (1 - cos2x) dx
= (1/2)∫ dx - (1/2)∫ cos2x dx
= (1/2)∫ dx - (1/2)(1/2)∫ cos2x d(2x)
= (1/2)x - (1/4)sin2x + C