Question

Evaluate each of the following integral:

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 x}{1 + e^x}dx\]

 

Answer 1

\[\text{Let I} =\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 x}{1 + e^x}dx................\left(1\right)\]

Then,

\[I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 \left[ \frac{\pi}{4} + \left( - \frac{\pi}{4} \right) - x \right]}{1 + e^\left[ \frac{\pi}{4} + \left( - \frac{\pi}{4} \right) - x \right]}dx .......................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 \left( - x \right)}{1 + e^{- x}}dx\]
\[ = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 x}{1 + \frac{1}{e^x}}dx\]
\[ = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{e^x \tan^2 x}{e^x + 1}dx . . . . . \left( 2 \right)\]

Adding (1) and (2), we get

\[2I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left( \frac{\tan^2 x}{1 + e^x} + \frac{e^x \tan^2 x}{1 + e^x} \right)dx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\left( 1 + e^x \right) \tan^2 x}{1 + e^x}dx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \tan^2 xdx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left( \sec^2 x - 1 \right)dx\]

\[\Rightarrow 2I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \sec^2 xdx - \int_{- \frac{\pi}{4}}^\frac{\pi}{4} dx\]
\[ \Rightarrow 2I = \tan x_{- \frac{\pi}{4}}^\frac{\pi}{4} - x_{- \frac{\pi}{4}}^\frac{\pi}{4} \]
\[ \Rightarrow 2I = \left[ \tan\frac{\pi}{4} - \tan\left( - \frac{\pi}{4} \right) \right] - \left[ \frac{\pi}{4} - \left( - \frac{\pi}{4} \right) \right]\]
\[ \Rightarrow 2I = \left( 1 + 1 \right) - \left( \frac{2\pi}{4} \right)\]
\[ \Rightarrow 2I = 2 - \frac{\pi}{2}\]
\[ \Rightarrow I = 1 - \frac{\pi}{4}\]