∫[t-sint)^2]sintdt =∫[t^2sint-2tsin^2(t)+sin^3(t)]dt=∫t^2sintdt-∫2tsin^2(t)dt+∫sin^3(t)dt
t取值[0,2π]时,积分sint、sin^3(t)均为0
故上式=-∫2tsin^2(t)dt=∫t[-2sin^2(t)]dt=∫t(cos2t-1)dt=-∫tdt+∫tcos2tdt(Ps:此时t取[0,2π])
=-2π^2+∫(t+π)cos2tdt(Ps:此时t取[-π,π])
=-2π^2+∫tcos2tdt++∫πcos2tdt
=-2π^2(PS:t取[-π,π]时,t为奇函数,cos2t为周期偶函数,∫tcos2tdt=0,∫cos2tdt=0)
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