Question Detail

This integral requires integration by parts, often 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, so du = cos x dx.Let dv = ex dx, so v = ex.I = ex sin x - ∫ ex cos x dxSecond application of integration by parts (for ∫ ex cos x dx):Let u = cos x, so du = -sin x dx.Let dv = ex dx, so v = ex.∫ ex cos x dx = ex cos x - ∫ ex (-sin x) dx∫ ex cos x dx = ex cos x + ∫ ex sin x dxNotice that ∫ ex sin x dx is our original integral I.So, ∫ ex cos x dx = ex cos x + I.Substitute this back into the first equation for I:I = ex sin x - (ex cos x + I)I = ex sin x - ex cos x - IAdd I to both sides:2I = ex sin x - ex cos x2I = ex (sin x - cos x)Divide by 2 and add the constant of integration:I = (1/2)ex (sin x - cos x) + CTherefore, the correct option is C.