QJEEM20Ma1a9d: JEE Main 2020 Mathematics - Class 12 JEEM20Ma1a9d Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx}

is:

Accepted Answer

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}}

Step-by-Step Verified Solution