答：
因?yàn)椤襵sin²x dx
=∫x(1-cos2x)/2 dx
=1/2∫x(1-cos2x) dx
=1/2∫x-xcos2x dx
=1/2(∫x dx - ∫xcos2x dx)
=x²/4-1/4xsin2x+1/4∫sin2x dx
=x²/4-1/4xsin2x-1/8cos2x + C
所以∫(0到1)xsin²x dx
=x²/4-1/4xsin2x-1/8cos2x |(0到1)
=1/4-sin2/4-cos2/8-(0-0-1/8)
=3/8-sin2/4-cos2/8