Question Detail

This integral requires integration by parts twice. Let I = ∫ ex sin x dx.First application of integration by parts (∫ u dv = uv - ∫ v du):Let u = sin x ⇒ du = cos x dxLet dv = ex dx ⇒ v = exSo, I = ex sin x - ∫ ex cos x dxNow, apply integration by parts again to the new integral ∫ ex cos x dx:Let u = cos x ⇒ du = -sin x dxLet dv = ex dx ⇒ v = exSo, ∫ ex cos x dx = ex cos x - ∫ ex (-sin x) dx = ex cos x + ∫ ex sin x dx.Notice that ∫ ex sin x dx is our original integral I. Substitute this back:∫ ex cos x dx = ex cos x + I.Now substitute this result 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)Finally, divide by 2 and add the constant of integration:I = (ex / 2) (sin x - cos x) + C.