Question

Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

Answer 1

We have

\[I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \frac{\cos x}{\left( 1 + e^x \right)}dx . . . . . \left( i \right)\]

\[\text{ Using property } \int_a^b f\left( x \right) dx = \int_a^b f\left( a + b - x \right) dx, \text { we get }\]

\[I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \frac{\cos\left( 0 - x \right)}{1 + e^\left( 0 - x \right)}dx\]

\[ = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \frac{\cos x}{1 + e^{- x}}dx\]

\[ \Rightarrow I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} e^x \frac{\left( \cos x \right)}{\left( 1 + e^x \right)}dx . . . . . \left( ii \right)\]

Adding (i) and (ii), we get

\[2I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \cos x dx = \left[ \sin x \right]_\frac{- \pi}{2}^\frac{\pi}{2} = 1 + 1 = 2\]

\[ \therefore I = 1\]