Question Detail

This is a classic cyclic integral that requires integration by parts twice. Let I = ∫ ex sin x dx. Apply integration by parts (∫ u dv = uv - ∫ v du) with u = sin x and dv = ex dx. Then du = cos x dx and v = ex. I = ex sin x - ∫ ex cos x dx. Now, apply integration by parts again to the new integral ∫ 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 + ∫ ex sin x dx. Notice that the new integral is our original integral, I. Substitute this back into the equation for I: I = ex sin x - (ex cos x + I). I = ex sin x - ex cos x - I. Now, solve for I: 2I = ex sin x - ex cos x. I = (ex⁄2) (sin x - cos x) + C. This matches option C.