Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the identity of Definite Integral that

$\displaystyle \int\limits_{- a}^{ a } f(x) \, dx = \displaystyle \int\limits_{0}^{ a } \bigg( f(x) + f( - x) \bigg)\, dx$

TO DETERMINE

$\displaystyle \int\limits_{- \frac{\pi}{2} }^{ \frac{\pi}{2} } \frac{1}{1 + {e}^{ \sin x} } \, dx$

EVALUATION

Using above mentioned Identity we get

$\displaystyle \int\limits_{- \frac{\pi}{2} }^{ \frac{\pi}{2} } \frac{1}{1 + {e}^{ \sin x} } \, dx$

$= \displaystyle \int\limits_{0 }^{ \frac{\pi}{2} } \bigg(\frac{1}{1 + {e}^{ \sin x} } + \frac{1}{1 + {e}^{ \sin ( - x )} } \bigg)\, dx$

$= \displaystyle \int\limits_{0 }^{ \frac{\pi}{2} } \bigg(\frac{1}{1 + {e}^{ \sin x} } + \frac{1}{1 + {e}^{ - \sin x} } \bigg)\, dx$

$= \displaystyle \int\limits_{0 }^{ \frac{\pi}{2} } \bigg(\frac{1}{1 + {e}^{ \sin x} } + \frac{1}{1 + \frac{1}{ {e}^{ \sin x}} } \bigg)\, dx$

$= \displaystyle \int\limits_{0 }^{ \frac{\pi}{2} } \bigg(\frac{1}{1 + {e}^{ \sin x} } + \frac{{e}^{ \sin x}}{1 + {e}^{ \sin x} } \bigg)\, dx$

$= \displaystyle \int\limits_{0 }^{ \frac{\pi}{2} } \frac{1 + {e}^{ \sin x}}{1 + {e}^{ \sin x} } \, dx$

$= \displaystyle \int\limits_{0 }^{ \frac{\pi}{2} } 1 \, dx$

$\displaystyle \: = \left x \right|_0^ \frac{\pi}{2}$

$\displaystyle \: = \frac{\pi}{2} - 0$

$\displaystyle \: = \frac{\pi}{2}$