Let $\displaystyle u = \cos{x}$, then $\displaystyle \dfrac{du}{dx} = -\sin{x} \Rightarrow {dx} = -\dfrac{du}{\sin{x}}. $
So $\displaystyle \int{\sin{x}(1+\cos{x})}\;{dx} = -\int\dfrac{\sin{x}(1+u)}{\sin{x}}\;{du} = -\int{\left(1+u\right)}\;{du} = -\left(u+\dfrac{u^2}{2}\right)+C$
Therefore $\displaystyle \int{\sin{x}(1+\cos{x})}\;{dx} = \boxed{-\dfrac{1}{2}\cos^2{x}-\cos{x}+C}.$