Question Detail

This integral requires applying integration by parts twice. Let I = ∫ eˣ cos x dx. For the first application, let u = cos x and dv = eˣ dx. Then du = -sin x dx and v = eˣ. So, I = eˣ cos x - ∫ eˣ (-sin x) dx = eˣ cos x + ∫ eˣ sin x dx. For the second integral (∫ eˣ sin x dx), again use integration by parts. Let u = sin x and dv = eˣ dx. Then du = cos x dx and v = eˣ. So, ∫ eˣ sin x dx = eˣ sin x - ∫ eˣ cos x dx = eˣ sin x - I. Substitute this back into the equation for I: I = eˣ cos x + (eˣ sin x - I). Now, solve for I: 2I = eˣ cos x + eˣ sin x. Therefore, I = (eˣ cos x + eˣ sin x)/2 + C.