Question

Evaluate each of the following integral:

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

Answer 1

\[\text{Let I} =\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos^2 x}{1 + e^x}dx.................\left(1\right)\]

Then,

\[I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos^2 \left[ \frac{\pi}{2} + \left( - \frac{\pi}{2} \right) - x \right]}{1 + e^\left[ \frac{\pi}{2} + \left( - \frac{\pi}{2} \right) - x \right]}dx ........................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos^2 \left( - x \right)}{1 + e^{- x}}dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{e^x \cos^2 x}{e^x + 1}dx .................... \left( 2 \right)\]
Adding (1) and (2), we get
\[2I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( \frac{\cos^2 x}{1 + e^x} + \frac{e^x \cos^2 x}{1 + e^x} \right)dx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos^2 x\left( 1 + e^x \right)}{1 + e^x}dx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos^2 xdx\]
\[ \Rightarrow 2I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( \frac{1 + \cos2x}{2} \right)dx\]

\[\Rightarrow 2I = \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} dx + \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos2xdx\]
\[ \Rightarrow 2I = \left.\frac{1}{2} \times x\right|_{- \frac{\pi}{2}}^\frac{\pi}{2} + \frac{1}{2} \left.\times \frac{\sin2x}{2}\right|_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ \Rightarrow 2I = \frac{1}{2}\left[ \frac{\pi}{2} - \left( - \frac{\pi}{2} \right) \right] + \frac{1}{4}\left[ \sin\pi - \sin\left( - \pi \right) \right]\]
\[ \Rightarrow 2I = \frac{1}{2} \times \pi + \frac{1}{4}\left( 0 + 0 \right) .....................\left[ \sin\left( - \pi \right) = - sin\pi = 0 \right]\]
\[ \Rightarrow 2I = \frac{\pi}{2}\]
\[ \Rightarrow I = \frac{\pi}{4}\]