Question Detail

This integral requires integration by parts twice. Let I = ∫ ex sinx dx.First application of integration by parts (∫ u dv = uv - ∫ v du):Let u = sinx, dv = ex dx.Then du = cosx dx, v = ex.I = ex sinx - ∫ ex cosx dx.Second application of integration by parts (to ∫ ex cosx dx):Let u' = cosx, dv' = ex dx.Then du' = -sinx dx, v' = ex.∫ ex cosx dx = ex cosx - ∫ ex (-sinx) dx = ex cosx + ∫ ex sinx dx.Notice that ∫ ex sinx dx is our original integral I.So, ∫ ex cosx dx = ex cosx + I.Substitute this back into the first equation for I:I = ex sinx - (ex cosx + I)I = ex sinx - ex cosx - IAdd I to both sides:2I = ex sinx - ex cosx2I = ex (sinx - cosx)I = (ex/2) (sinx - cosx) + C.This matches option C.