$I=\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$
$$
\begin{aligned}
I &=\int_{-\pi / 2}^{\pi / 2} \frac{\cos (\pi / 2-\pi / 2-x)}{1+e^{(\pi / 2-\pi / 2-x)}} d x \\
&=\int_{-\pi / 2}^{\pi / 2} \frac{\cos (-x)}{1+e^{-x}} d x \\
I &=\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{-x}} d x \\
&=\int_{-\pi / 2}^{\pi / 2} \frac{e^{x} \cos x}{1+e^{x}} d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I &=\int_{-\pi / 2}^{\pi / 2} \frac{\left(1+e^{x}\right) \cos x}{\left(1+e^{x}\right)} d x \\
&=\int_{-\pi / 2}^{\pi / 2} \cos x d x \\
&=2 \int_{0}^{\pi / 2} \cos x d x
\end{aligned}
$$
[Since, $\cos x$ is an even function.]
$$
\therefore \quad 2 I=2[\sin x]_{0}^{\pi / 2}=2(1-0)=2
$$
$$
\Rightarrow \quad I=1
$$