Question Detail

This integral requires integration by parts twice, as it is a cyclic integral. Let I = ∫ex cos x dx.For the first application of integration by parts, let u = cos x and dv = ex dx.Then du = -sin x dx and v = ex.I = ex cos x - ∫ex(-sin x) dxI = ex cos x + ∫ex sin x dxNow, apply integration by parts to ∫ex sin x dx. Let u = sin x and dv = ex dx.Then du = cos x dx and v = ex.∫ex sin x dx = ex sin x - ∫ex cos x dxSubstitute this back into the expression for I:I = ex cos x + (ex sin x - ∫ex cos x dx)I = ex cos x + ex sin x - INow, move the -I term to the left side:2I = ex cos x + ex sin xI = (ex cos x + ex sin x) / 2 + CI = ex(cos x + sin x) / 2 + CThis matches option B.