DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

$\int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\text{dx}=\int\limits_{\text{a}}^{\text{b}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$

Hence,

$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{(-x)}}{1+\text{e}^\text{-x}}\text{dx}$

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

If,

$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

Then

$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

So,

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{(1+\text{e}^\text{x})\cos^2\text{x}}{1+\text{e}^\text{x}}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\text{x}\text{dx}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$

$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{4}\bigg\{\text{x}+\frac{\sin2\text{x}}{2}\bigg\}^\frac{\pi}{2}_{-\frac{\pi}{2}}$

$\text{I}=\frac{1}{4}\bigg\{\bigg(\frac{\pi}{2}\bigg)-\bigg(-\frac{\pi}{2}\bigg)\bigg\} $

$\text{I}=\frac{\pi}{4}$