Evaluate the following integrals: ^ 3 _ - 3 1 1+e^ dx - Vidyadip

Question

Evaluate the following integrals:$\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}$

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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\big[\frac{\pi}{3}+\big(-\frac{\pi}{3}\big)-\text{x}\big]}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan(-\text{x})}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{-\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{\tan\text{x}}}{\text{e}^{\tan\text{x}}+1}\text{ dx}\ ...{\text{(ii)}}$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{1+\tan\text{x}}}{\text{e}^{1+\tan\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_{-\frac{\pi}{3}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\Big(-\frac{\pi}{3}\Big)$
$\Rightarrow2\text{I}=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$

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