Question

Evaluate each of the following integrals:

${\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+e^{\tan x}}dx$

Answer 1

Let I = ${\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+e^{\tan x}}dx$ .....(1)
Then,
$$\begin{gathered} I={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+e^{\tan \left[{\displaystyle \frac{\mathrm{\pi }}{3}}+\left(-{\displaystyle \frac{\mathrm{\pi }}{3}}\right)-x\right]}}dx\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ ={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+e^{\tan \left(-x\right)}}dx \\ ={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+e^{-\tan x}}dx \\ ={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{e^{\tan x}}{e^{\tan x}+1}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\frac{1+e^{\tan x}}{1+e^{\tan x}}dx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}dx \\ \Rightarrow 2I={\overline{)x}}_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}} \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{3}-\left(-\frac{\mathrm{\pi }}{3}\right)=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow I=\frac{\mathrm{\pi }}{3} \end{gathered}$$