\(\int_{-\pi / 2}^{-\pi / 2} \frac{\cos x}{1+e^{x}} d x\) is equal to

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