Question

Evaluate each of the following integral:

\[\int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^\ tan\ x}dx\]

 

Answer 1

\[\text{Let I} =\int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^\ tan\ x}dx.................\left(1\right)\]

Then,

\[I = \int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^\ tan\left[ \frac{\pi}{3} + \left( - \frac{\pi}{3} \right) - x \right]}dx ..................\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ = \int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^{\ tan}\left( - x \right)}dx\]
\[ = \int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^{{- \ tan x}}}dx\]
\[ = \int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{e^{\ tan} x}{e^{\ tan} x + 1}dx . . . . . \left( 2 \right)\]

Adding (1) and (2), we get

\[2I = \int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1 + e^{\ tan x}}{1 + e^{\ tan x}}dx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{3}}^\frac{\pi}{3} dx\]
\[ \Rightarrow 2I = \left.x\right|_{- \frac{\pi}{3}}^\frac{\pi}{3} \]
\[ \Rightarrow 2I = \frac{\pi}{3} - \left( - \frac{\pi}{3} \right) = \frac{2\pi}{3}\]
\[ \Rightarrow I = \frac{\pi}{3}\]