Given that $y = e^x (1+\cos x)$, show $\dfrac{dy}{dx} = e^x (1+\cos x - \sin x)$. Hence find $\displaystyle \int e^x (\cos x - \sin x)\,dx$.

1. Apply the product rule to differentiate the function

Let $u = e^x$ and $v = 1+\cos x$.
$\dfrac{du}{dx} = e^x$
$\dfrac{dv}{dx} = -\sin x$
$\dfrac{dy}{dx} = \dfrac{du}{dx}v + u\dfrac{dv}{dx}$
$\dfrac{dy}{dx} = e^x(1+\cos x) + e^x(-\sin x)$

2. Factorise to show the required result

$\dfrac{dy}{dx} = e^x(1+\cos x - \sin x)$

1. Express the integral of the derivative from part (a)

From (a), $\frac{d}{dx}[e^x(1+\cos x)] = e^x(1+\cos x - \sin x)$.
Integrating both sides with respect to $x$:
$\int e^x(1+\cos x - \sin x)\,dx = e^x(1+\cos x) + c_1$

2. Split the integral to isolate the required term

$\int (e^x + e^x(\cos x - \sin x))\,dx = e^x(1+\cos x) + c_1$
$\int e^x\,dx + \int e^x(\cos x - \sin x)\,dx = e^x(1+\cos x) + c_1$

3. Evaluate the standard integral and rearrange to find the final answer

$e^x + \int e^x(\cos x - \sin x)\,dx = e^x(1+\cos x) + c_1$
$\int e^x(\cos x - \sin x)\,dx = e^x(1+\cos x) - e^x + c$