✔ 最佳答案
sin t cos nt = [sin (n + 1)t + sin (1 - n)t]/2
= [sin (n + 1)t - sin (n - 1)t]/2
∫sin t cos nt dt = ∫ {[sin (n + 1)t - sin (n - 1)t]/2} dt
= cos (n - 1)t/[2(n - 1)] - cos (n + 1)t/[2(n + 1)] + C
2010-02-08 17:12:14 補充:
n 為 1 時:
∫sin t cos nt dt = ∫sin t cos t dt
= (1/2)∫sin 2t dt
= -(cos 4t)/2 + C
n 不是 1 時:
∫sin t cos nt dt = cos (n - 1)t/[2(n - 1)] - cos (n + 1)t/[2(n + 1)] + C