The value of $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ is:
Solution
I = $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ ....(1)
<br><br>Replacing x with $\left( {{\pi \over 2} - {\pi \over 2} + x} \right)$, we get
<br><br>I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{dx} \over {1 + {e^{ - \sin x}}}}} $$
<br><br>= $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{e^{\sin x}}dx} \over {1 + {e^{\sin x}}}}} $$ .....(2)
<br><br>Adding (1) and (2), we get
<br><br>2I = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {1dx}$ = $\left( {{\pi \over 2} - \left( { - {\pi \over 2}} \right)} \right)$ = $\pi$
<br><br>$\Rightarrow$ I = ${{\pi \over 2}}$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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