[tex]\displaystyle I = \int_{0}^{\frac{\pi}{2}}\Big(\dfrac{1+\sin x}{1+\cos x}\cdot e^{x}\, dx\Big)=\\ \\ =\int_{0}^{\frac{\pi}{2}} e^x \cdot \dfrac{\sin x}{1+\cos x}\, dx +\int_{0}^{\frac{\pi}{2}}e^x\cdot \dfrac{1}{1+\cos x}\, dx = \\ \\ = \int_{0}^{\frac{\pi}{2}} e^x \cdot \dfrac{\sin x}{1+\cos x}\, dx+\int_{0}^{\frac{\pi}{2}} e^x \cdot \Big(\dfrac{\sin x}{1+\cos x}\Big)'\, dx=[/tex]
[tex]\displaystyle = \int_{0}^{\frac{\pi}{2}} e^x \cdot \dfrac{\sin x}{1+\cos x}\, dx+\\ +e^x \cdot \dfrac{\sin x}{1+\cos x}\Big|_{0}^{\frac{\pi}{2}}- \int_{0}^{\frac{\pi}{2}} e^x \cdot \dfrac{\sin x}{1+\cos x}\, dx = \\ \\ \\= e^x \cdot \dfrac{\sin x}{1+\cos x}\Big|_{0}^{\frac{\pi}{2}} = \\ \\ = e^{\frac{\pi}{2}}\cdot 1 - 0 = e^{\frac{\pi}{2}} \\ \\ \Rightarrow \boxed{k = \dfrac{\pi}{2}}[/tex]
