Challenging Integral

Relevant wiki: Integration Tricks

Similar solution as @Jam M's

$\begin{aligned} I & = \int_{-\frac \pi 2}^\frac \pi 2 \frac {\cos x}{e^\frac 1x + 1}dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 \left(\frac {\cos x}{e^\frac 1x + 1} + \frac {\color{#3D99F6} \cos (-x)}{e^{-\frac 1x} + 1} \right) dx & \small \color{#3D99F6} \text{Note that }\cos (-\theta) = \cos \theta \\ & = \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 \cos x \left(\frac 1{e^\frac 1x + 1} + \frac {e^\frac 1x}{1 + e^\frac 1x} \right) dx \\ & = \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 \cos x \ dx \\ & = \frac {\sin x}2 \bigg|_{-\frac \pi 2}^\frac \pi 2 \\ & = \boxed 1 \end{aligned}$

Any function $f(x)$ can be expressed as $f(x) = f_e(x) + f_o(x)$ , where $f_e$ and $f_o$ are even and odd functions respectively.

To see this, note that:

$f_e(x) = \dfrac{f(x) + f(-x)}{2}$ is even, $f_o(x) = \dfrac{f(x) - f(-x)}{2}$ is odd, and $f(x) = f_e(x) + f_o(x)$ .

Let $f(x) = \dfrac{\cos x}{e^{1/x} + 1}$ so that

$\displaystyle{\int_{-\pi/2}^{\pi/2} \dfrac{\cos x}{e^{1/x} + 1} dx} = \displaystyle{\int_{-\pi/2}^{\pi/2} f_e(x) dx} + \displaystyle{\int_{-\pi/2}^{\pi/2} f_o(x) dx}$

Since $f_o$ is odd, $\displaystyle{\int_{-\pi/2}^{\pi/2} f_o(x) dx} = 0$ . Furthermore,

$f_e(x) = \dfrac{1}{2} \left( \dfrac{\cos x}{e^{1/x} + 1} + \dfrac{\cos(-x)}{e^{-1/x} + 1} \right) = \dfrac{\cos x}{2} \left( \dfrac{e^{-1/x} + 1 + e^{1/x} + 1}{1 + e^{1/x} + e^{-1/x} + 1} \right) = \dfrac{\cos x}{2}$

$\displaystyle{\int_{-\pi/2}^{\pi/2} \dfrac{\cos x}{e^{1/x} + 1} dx} = \displaystyle{\int_{-\pi/2}^{\pi/2} \dfrac{\cos x}{2} dx} = \dfrac{\sin x}{2} \Big|_{-\pi/2}^{\pi/2} = 1$