Question Detail

This integral requires integration by parts twice. The formula for integration by parts is ∫ u dv = uv - ∫ v du.Let I = ∫ ex sin x dx.First application of integration by parts:Let u = sin x and dv = ex dx.Then du = cos x dx and v = ex.I = ex sin x - ∫ ex cos x dx (Equation 1)Second application of integration by parts (on ∫ ex cos x dx):Let u = cos x and dv = ex dx.Then du = -sin x dx and v = ex.∫ ex cos x dx = ex cos x - ∫ ex (-sin x) dx∫ ex cos x dx = ex cos x + ∫ ex sin x dx (Equation 2)Substitute Equation 2 back into Equation 1:I = ex sin x - (ex cos x + ∫ ex sin x dx)I = ex sin x - ex cos x - IAdd I to both sides:2I = ex sin x - ex cos xI = (ex/2)(sin x - cos x) + C.This matches option C.