JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 17)
The value of $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ is:
$$\pi $$
$${{3\pi \over 2}}$$
$${{\pi \over 2}}$$
$${{\pi \over 4}}$$
Explanation
I = $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ ....(1)
Replacing x with $$\left( {{\pi \over 2} - {\pi \over 2} + x} \right)$$, we get
I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{dx} \over {1 + {e^{ - \sin x}}}}} $$
= $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{e^{\sin x}}dx} \over {1 + {e^{\sin x}}}}} $$ .....(2)
Adding (1) and (2), we get
2I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {1dx} $$ = $$\left( {{\pi \over 2} - \left( { - {\pi \over 2}} \right)} \right)$$ = $$\pi $$
$$ \Rightarrow $$ I = $${{\pi \over 2}}$$
Replacing x with $$\left( {{\pi \over 2} - {\pi \over 2} + x} \right)$$, we get
I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{dx} \over {1 + {e^{ - \sin x}}}}} $$
= $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{e^{\sin x}}dx} \over {1 + {e^{\sin x}}}}} $$ .....(2)
Adding (1) and (2), we get
2I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {1dx} $$ = $$\left( {{\pi \over 2} - \left( { - {\pi \over 2}} \right)} \right)$$ = $$\pi $$
$$ \Rightarrow $$ I = $${{\pi \over 2}}$$
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