Evaluate the following: ^1_0 dx e^x+e^ -x dx - Vidyadip

Question

Evaluate the following:
$\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}\text{dx}$

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Answer

Let $\text{I}=\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
$=\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\frac{1}{\text{e}^\text{x}}}$
$=\int\limits^0_1\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}$
Put $\text{e}^\text{x}=\text{t}$
$\text{e}^\text{x}\text{dx}=\text{dt}$
Substituting ex = t and ex dx = dt
$\therefore\ \text{I}=\int\limits^\text{e}_1\frac{\text{dt}}{1+\text{t}^2}$
$=\big[\tan^{-1}\text{t}\big]^\text{e}_\text{1}$ $\Big[\because\int\frac{1}{1+\text{x}^2}\text{dx}=\tan^{-1}\text{x}+\text{C}\Big]$
$=\tan^{-1}\text{e}-\tan^{-1}1$
$=\tan^{-1}\text{e}-\frac{\pi}{4}$ $\Big[\because\tan^{-1}1=\frac{\pi}{4}\Big]$

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