Question

求该题分部积分过程

Answer 1

∫ e^[(R/L)t ]. sinωt dt
=(L/R)∫ sinωt de^[(R/L)t ]
=(L/R)sinωt.e^[(R/L)t ] -(Lω/R)∫ cosωt .e^[(R/L)t dt
=(L/R)sinωt.e^[(R/L)t ] -(L^2.ω/R^2)∫ cosωt de^[(R/L)t]
=(L/R)sinωt.e^[(R/L)t ] -(L^2.ω/R^2).cosωt.e^[(R/L)t ]
- (L^2.ω^2/R^2)∫ sinωt .e^[(R/L)t ] dt
[(R^2+L^2.ω^2)/R^2]∫ e^[(R/L)t ]. sinωt dt
=(L/R)sinωt.e^[(R/L)t ] -(L^2.ω/R^2).cosωt.e^[(R/L)t ]
∫ e^[(R/L)t ]. sinωt dt
=[R^2/(R^2+L^2.ω^2) ] { (L/R)sinωt.e^[(R/L)t ] -(L^2.ω/R^2).cosωt.e^[(R/L)t ] }
=[1/(R^2+L^2.ω^2) ] { LRsinωt.e^[(R/L)t ] - L^2.ω.cosωt.e^[(R/L)t ] }
={ e^[(R/L)t ] /(R^2+L^2.ω^2) } { LRsinωt - L^2.ω.cosωt }

Answer 2

如图所示