赞 zhwsh3333 春芽
共回答了17个问题采纳率：94.1% 举报
用积化和差公式：sinxsiny=(1/2)[cos[(x-y)-cos(x+y)],cosxcosy=(1/2)[cos(x+y)+cos(x-y)]
∫sint*sinωt dt
= (1/2)∫cos(t-ωt) dt - (1/2)∫cos(t+ωt) dt
= (1/2)∫cos(1-ω)t dt - (1/2)∫cos(1+ω)t dt
= (1/2)/(1-ω)∫cos(1-ω)t d(1-ω)t - (1/2)/(1+ω)∫cos(1+ω)t d(1+ω)t
= 1/[2(1-ω)] * sin[(1-ω)t] - 1/[2(1+ω)] * sin[(1+ω)t] + C
∫cost*cosωt dt
= (1/2)∫cos(t+ωt) dt + (1/2)∫cos(t-ωt) dr
= (1/2)/(1+ω)∫cos(1+ω)t d(1+ω)t + (1/2)/(1-ω)∫cos(1-ω)t d(1-ω)t
= 1/[2(1+ω)] * sin[(1+ω)t] + 1/[2(1-ω)] * sin[(1-ω)t] + C
∫sint*sinωt dt
= (1/2)∫cos(t-ωt) dt - (1/2)∫cos(t+ωt) dt
= (1/2)∫cos(1-ω)t dt - (1/2)∫cos(1+ω)t dt
= (1/2)/(1-ω)∫cos(1-ω)t d(1-ω)t - (1/2)/(1+ω)∫cos(1+ω)t d(1+ω)t
= 1/[2(1-ω)] * sin[(1-ω)t] - 1/[2(1+ω)] * sin[(1+ω)t] + C
∫cost*cosωt dt
= (1/2)∫cos(t+ωt) dt + (1/2)∫cos(t-ωt) dr
= (1/2)/(1+ω)∫cos(1+ω)t d(1+ω)t + (1/2)/(1-ω)∫cos(1-ω)t d(1-ω)t
= 1/[2(1+ω)] * sin[(1+ω)t] + 1/[2(1-ω)] * sin[(1-ω)t] + C
1年前
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