Evaluate the following integrals: ^ a _ -a 1 1+a^ x dx, a>0 - Vidyadip

Question

Evaluate the following integrals:$\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{\text{x}}}\text{ dx},\text{ a}>0$

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Answer

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

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