Evaluate: `int_(-pi//4)^(pi//4)(tan^2x)/(1+e^x)dx`

Explanation

`I=int_{-frac{x}{4}}^{frac{x}{4}} frac{tan ^{2} x}{1+e^{x}} d x………….(1)`

`

I=int_{-frac{x}{4}}^{frac{x}{4}} frac{tan ^{2}[frac{pi}{4}+(-frac{pi}{4})-x]}{1+e^{[frac{x}{x}+(-frac{4}{4})-x]}} d x ldots ldots . ldots ldots . ldots . ldots[int_{a}^{b} f(x) d x=int_{a}^{b} f(a+b-x) d x] `

`
=int_{-frac{x}{4}}^{frac{x}{4}} frac{tan ^{2}(-x)}{1+e^{-x}} d x`

`
=int_{-frac{x}{4}}^{frac{pi}{4}} frac{e^{x} tan ^{2} x}{e^{x}+1} d x ……………………(2) `


By Adding (1) and (2), we obtain

`
2 I=int_{-frac{x}{4}}^{frac{x}{4}}(frac{tan ^{2} x}{1+e^{x}}+frac{e^{x} tan ^{2} x}{1+e^{x}}) d x`

`
Rightarrow 2 I=int_{-frac{x}{4}}^{frac{x}{4}} frac{(1+e^{x}) tan ^{2} x}{1+e^{x}} d x`

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