Definite Integrals (1)

$I = \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \dfrac{\cos x}{1+e^{x}} dx$
$I =\displaystyle \int_{0}^{\frac{\pi}{2}}\dfrac{\cos x}{1+e^{x}} +\dfrac{\cos x}{1+e^{-x}} dx = \displaystyle \int_{0}^{\frac{\pi}{2}} \cos x dx$

$I = \displaystyle \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = 1 - 0 = 1$

$\displaystyle I = \int_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx \\ \text{Using property of definite integral, } \\ I = \int_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \frac{\cos{(-x)}}{1+e^{-x}} dx \\ = \int_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \frac{e^x\cos{x}}{1+e^{x}} dx \\ \therefore 2I = \int_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \cos x \ dx \\ \therefore I = \frac{1}{2} \sin x \Bigr|_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \\ = \frac{1+1}{1} = \boxed{1}$